### 12.2. Controlled and conditioned invariance

We have denoted $(\mathcal{V}_W)_{\dot{W}}$ by $\mathcal{V}_{\dot{W}}$ in this subsection.

**Definition 38.** We say $\mathcal{V}_W$ is controlled invariant in $\mathcal{V}_{W\dot{W}M}$ relative to $\mathcal{V}_M$ or controlled invariant in $\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M$ iff

$$ (\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M) \leftrightarrow \mathcal{V}_{\dot{W}} \supseteq \mathcal{V}_W. \qquad (49) $$

We say $\mathcal{V}_W$ is conditioned invariant in $\mathcal{V}_{W\dot{W}M}$ relative to $\mathcal{V}_M$ or conditioned invariant in $\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M$ iff

$$ (\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M) \leftrightarrow \mathcal{V}_W \subseteq \mathcal{V}_{\dot{W}}. \qquad (50) $$

**Problem 43.** For a system defined in the state-input-output form $\dot{w} = Aw + Bu, y = Cw$, show that $\mathcal{V}_W$ is controlled invariant iff $A(\mathcal{V}_W) \subseteq \mathcal{V}_W + \mathcal{B}$, where $\mathcal{B} = \text{Im}(B)$.

**Answer**

We say $\mathcal{V}_W$ is controlled invariant in $\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M$ iff $(\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M) \leftrightarrow \mathcal{V}_{\dot{W}} \supseteq \mathcal{V}_W$. In the present case we take $\mathcal{V}_M$ to be $\mathcal{F}_U \oplus \mathcal{F}_Y$. The condition for controlled invariance is equivalent to the statement 'whenever $w \in \mathcal{V}_W$, there exists a vector $b \in \mathcal{B}$ and a vector $\dot{w} \in \mathcal{V}_{\dot{W}}$ such that the vector $(\dot{w}, w, u)$, where $b = Bu$, satisfies $\dot{w} = Aw + Bu$. This is clearly equivalent to the statement 'whenever $w \in \mathcal{V}_W$, we can find a $b \in \mathcal{B}$ such that $Aw + b$ lies in $\mathcal{V}_W$, i.e., equivalent to $A(\mathcal{V}_W) \subseteq \mathcal{V}_W + \mathcal{B}$.'

**Problem 44.** For a system defined in the state-input-output form $\dot{w} = Aw + Bu, y = Cw$, show that $\mathcal{V}_W$ is conditioned invariant iff $A(\mathcal{V}_W \cap \mathcal{C}) \subseteq \mathcal{V}_W$, where $\mathcal{C} = \text{Ker}(C)$.

**Answer**

We say $\mathcal{V}_W$ is conditioned invariant in $\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M$ iff $(\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M) \leftrightarrow \mathcal{V}_W \subseteq \mathcal{V}_{\dot{W}}$. In the present case we take $\mathcal{V}_M$ to be $\mathbf{0}_U \oplus \mathbf{0}_Y$. The statement about conditioned invariance is equivalent to the statement 'whenever $w \in \mathcal{V}_W$, if $y = 0$, i.e., if $Cw = 0$, then the vector $\dot{w} = Aw$, lies in $\mathcal{V}_{\dot{W}}$. This is clearly equivalent to the statement 'whenever $w \in \mathcal{V}_W \cap \mathcal{C}$, $Aw \in \mathcal{V}_W$. i.e., equivalent to $A(\mathcal{V}_W \cap \mathcal{C}) \subseteq \mathcal{V}_W$.'

### 12.3. Minimal conditioned invariance and maximal controlled invariance

**Problem 45.** Let $\mathcal{V}_{W\dot{W}UY}$ be a dynamical system defined as in Equation 44 and let $\mathcal{V}_{W\dot{W}} := \mathcal{V}_{W\dot{W}UY} \leftrightarrow \mathcal{V}_{UY}$. Let $\mathcal{B}_W, \mathcal{C}_W$ be subspaces on $W$. Show that there is a unique minimal conditioned invariant space of $\mathcal{V}_{W\dot{W}}$ containing $\mathcal{B}_W$ and a unique maximal controlled invariant space of $\mathcal{V}_{W\dot{W}}$ contained in $\mathcal{C}_W$ by justifying the following algorithms ([12]).

#### Algorithm I

Algorithm for computation of unique minimal space containing $\mathcal{B}_W$ and satisfying the condition $(\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M) \leftrightarrow \mathcal{V}_W \subseteq \mathcal{V}_{\dot{W}}$

Let us denote $(\mathcal{V}_{W\dot{W}M} \leftrightarrow \mathcal{V}_M)$ by $\mathcal{V}_{W\dot{W}}$. Let $\mathcal{V}_W^1 = \mathcal{B}_W$.

1. Check if $\mathcal{V}_{W\dot{W}} \leftrightarrow \mathcal{V}_W^1 \subseteq \mathcal{V}_{\dot{W}}^1$. If so, this is the desired subspace.

2. If not, let $\mathcal{V}_{\dot{W}}^2 = (\mathcal{V}_{W\dot{W}} \leftrightarrow \mathcal{V}_W^1) + \mathcal{V}_{\dot{W}}^1$ and $\mathcal{V}_W^2$ be the copy of $\mathcal{V}_{\dot{W}}^2$ on $W$. (In this case observe that $\dim(\mathcal{V}_W^2) > \dim(\mathcal{V}_W^1)$).
3. Repeat taking $\mathcal{Y}_W^{j+1}$ to be the $W$-copy of $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^j) + \mathcal{Y}_{\dot{W}}^j$ and stop when $\mathcal{Y}_W^{n+1} = \mathcal{Y}_W^n$. $\mathcal{Y}_W^n$ is the desired minimal space.

**Algorithm II**

Algorithm for computation of unique maximal space contained in $C_W$ and satisfying the condition
($\mathcal{Y}_{W\dot{W}M} \leftrightarrow \mathcal{Y}_M) \leftrightarrow \mathcal{Y}_{\dot{W}} \supseteq \mathcal{Y}_W$.

Let us denote $(\mathcal{Y}_{W\dot{W}M} \leftrightarrow \mathcal{Y}_M)$ by $\mathcal{Y}_{W\dot{W}}$. Let $\mathcal{Y}_W^1 = C_W$.

1. Check if $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^1 \supseteq \mathcal{Y}_W^1$. If so this is the desired subspace.

2. If not let $\mathcal{Y}_W^2 = (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^1) \cap \mathcal{Y}_W^1$ and $\mathcal{Y}_{\dot{W}}^2$ be the copy of $\mathcal{Y}_W^2$ on $\dot{W}$. In this case observe that $\dim(\mathcal{Y}_W^2) < \dim(\mathcal{Y}_W^1)$.

3. Repeat taking $\mathcal{Y}_{\dot{W}}^{j+1}$ to be the $\dot{W}$-copy of $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^j) \cap \mathcal{Y}_W^j$ and stop when $\mathcal{Y}_W^{n+1} = \mathcal{Y}_W^n$. $\mathcal{Y}_W^n$ is the desired maximal space.

Answer

Justification for Algorithm I

We have $\mathcal{Y}_W^1 = \mathcal{B}_W$, $\mathcal{Y}_W^{j+1} \supseteq \mathcal{Y}_W^j$ and $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) + \mathcal{Y}_{\dot{W}}^n = \mathcal{Y}_{\dot{W}}^{n+1} = \mathcal{Y}_{\dot{W}}^n$. Thus $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) + \mathcal{Y}_{\dot{W}}^n \subseteq \mathcal{Y}_{\dot{W}}^n$ and therefore, $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) \subseteq \mathcal{Y}_{\dot{W}}^n$. Hence $\mathcal{Y}_{\dot{W}}^n$ is conditioned invariant and contains $\mathcal{B}_W$.

Next, observe that if $\hat{\mathcal{Y}}_W \supseteq \mathcal{Y}_W^j$, then we have $\hat{\mathcal{Y}}_{\dot{W}} \supseteq \mathcal{Y}_{\dot{W}}^j$ and further, if $\hat{\mathcal{Y}}_W$ is conditioned invariant
then $\hat{\mathcal{Y}}_{\dot{W}} \supseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \hat{\mathcal{Y}}_W) \supseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^j)$. Hence $\hat{\mathcal{Y}}_{\dot{W}} \supseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^j) + \mathcal{Y}_{\dot{W}}^j \supseteq \mathcal{Y}_{\dot{W}}^{j+1}$. It follows that
if $\hat{\mathcal{Y}}_W \supseteq \mathcal{B}_W$ and $\hat{\mathcal{Y}}_W$ is conditioned invariant, we must have $\hat{\mathcal{Y}}_W$ containing all $\mathcal{Y}_{\dot{W}}^j$ and therefore also
$\mathcal{Y}_W^{n+1} = \mathcal{Y}_W^n$. It thus follows that $\mathcal{Y}_W^n$ is the unique minimal subspace containing $\mathcal{B}_W$ that is conditioned
invariant.

Justification for Algorithm II

We have $\mathcal{Y}_W^1 = C_W$, $\mathcal{Y}_W^{j+1} \subseteq \mathcal{Y}_W^j$ and $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) \cap \mathcal{Y}_W^n = \mathcal{Y}_W^{n+1} = \mathcal{Y}_W^n$. Thus $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) \cap \mathcal{Y}_W^n \supseteq \mathcal{Y}_W^n$ and therefore $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^n) \supseteq \mathcal{Y}_W^n$. Hence $\mathcal{Y}_W^n$ is controlled invariant and is contained in $C_W$.

Next, observe that if $\hat{\mathcal{Y}}_W \subseteq \mathcal{Y}_W^j$, then $(\mathcal{Y}_{W\dot{W}} \leftrightarrow \hat{\mathcal{Y}}_W) \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W^j)$. Now if $\hat{\mathcal{Y}}_W$ is controlled invariant,
then $\hat{\mathcal{Y}}_W \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \hat{\mathcal{Y}}_{\dot{W}}) \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^j)$. Therefore, since $\hat{\mathcal{Y}}_W \subseteq \mathcal{Y}_W^j$, $\hat{\mathcal{Y}}_W \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^j) \cap \mathcal{Y}_W^j = \mathcal{Y}_{\dot{W}}^{j+1}$.
It follows that if $\hat{\mathcal{Y}}_W \subseteq C_W$ and $\hat{\mathcal{Y}}_W$ is controlled invariant then we must have $\hat{\mathcal{Y}}_W$ contained in all $\mathcal{Y}_{\dot{W}}^j$ and therefore, also $\mathcal{Y}_W^n = \mathcal{Y}_W^{n+1}$. It thus follows that $\mathcal{Y}_W^n$ is the unique maximal controlled invariant subspace contained in $C_W$.

**13. Duality**

13.1. Vector space duality

**Problem 46.** The duals of the following 'primal' statements are given in the succeeding paragraph by interchanging $\subseteq, \supseteq; +, \cap; \circ, \times; \leftrightarrow, \Leftrightarrow$. (Note that $(\mathcal{Y}_{SP} \Leftrightarrow \mathcal{Y}_P) = (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P)$.) Assuming that the primal statements are true, prove the duals.

Primal

1. $(\mathcal{Y}_{SP}^1 + \mathcal{Y}_{SP}^2) \leftrightarrow \mathcal{Y}_P \supseteq (\mathcal{Y}_{SP}^1 \leftrightarrow \mathcal{Y}_P) + (\mathcal{Y}_{SP}^2 \leftrightarrow \mathcal{Y}_P)$.

2. $\mathcal{Y}_{SP} \leftrightarrow (\mathcal{Y}_P^1 \cap \mathcal{Y}_P^2) \subseteq (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P^1) \cap (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P^2)$.
3. $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = \mathcal{Y}_{SP} \leftrightarrow (\mathcal{Y}_{SP} \circ P \cap \mathcal{Y}_P)$.

4. $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = (\mathcal{Y}_{SP} \cap (\mathcal{Y}_{SP} \circ S \oplus \mathcal{Y}_P)) \leftrightarrow \mathcal{Y}_P$.

5. Let $\mathcal{Y}_{SP}, \mathcal{Y}_S$ be vector spaces on $S \uplus P, S$, respectively. Then there exists a vector space $\mathcal{Y}_P$ such that $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = \mathcal{Y}_S$ iff $\mathcal{Y}_{SP} \times S \subseteq \mathcal{Y}_S \subseteq \mathcal{Y}_{SP} \circ S$.

### Dual

1. $(\mathcal{Y}_{SP}^1 \cap \mathcal{Y}_{SP}^2) \leftrightarrow \mathcal{Y}_P \subseteq (\mathcal{Y}_{SP}^1 \leftrightarrow \mathcal{Y}_P) \cap (\mathcal{Y}_{SP}^2 \leftrightarrow \mathcal{Y}_P)$.

2. $\mathcal{Y}_{SP} \leftrightarrow (\mathcal{Y}_P^1 + \mathcal{Y}_P^2) \supseteq (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P^1) + (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P^2)$.

3. $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = \mathcal{Y}_{SP} \leftrightarrow (\mathcal{Y}_{SP} \times P + \mathcal{Y}_P)$.

4. $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = (\mathcal{Y}_{SP} + (\mathcal{Y}_{SP} \times S \oplus \mathcal{Y}_P)) \leftrightarrow \mathcal{Y}_P$.

5. (This statement is self dual.) Let $\mathcal{Y}_{SP}, \mathcal{Y}_S$ be vector spaces on $S \uplus P, S$, respectively. Then there exists a vector space $\mathcal{Y}_P$ such that $\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P = \mathcal{Y}_S$ iff $\mathcal{Y}_{SP} \times S \subseteq \mathcal{Y}_S \subseteq \mathcal{Y}_{SP} \circ S$.

### Answer

We prove only the dual statement in (1) in detail. The proofs of remaining statements are similar.

*Proof.* We have $((\mathcal{Y}_{SP}^1)^\perp + (\mathcal{Y}_{SP}^2)^\perp) \leftrightarrow \mathcal{Y}_P^\perp \supseteq ((\mathcal{Y}_{SP}^1)^\perp \leftrightarrow \mathcal{Y}_P^\perp) + ((\mathcal{Y}_{SP}^2)^\perp \leftrightarrow \mathcal{Y}_P^\perp)$.

Therefore $[(\mathcal{Y}_{SP}^1)^\perp + (\mathcal{Y}_{SP}^2)^\perp) \leftrightarrow \mathcal{Y}_P^\perp] \supseteq [(\mathcal{Y}_{SP}^1)^\perp \leftrightarrow \mathcal{Y}_P^\perp) + ((\mathcal{Y}_{SP}^2)^\perp \leftrightarrow \mathcal{Y}_P^\perp)]^\perp$. By

Theorem 2, we have $((\mathcal{Y}_{SP}^1)^\perp + (\mathcal{Y}_{SP}^2)^\perp)^\perp \leftrightarrow (\mathcal{Y}_P^\perp)^\perp \subseteq [(\mathcal{Y}_{SP}^1)^\perp \leftrightarrow \mathcal{Y}_P^\perp]^\perp \cap [(\mathcal{Y}_{SP}^2)^\perp \leftrightarrow \mathcal{Y}_P^\perp]^\perp$.

Using Theorem 2, the duality between sum and intersection and the fact that $((\mathcal{Y})^\perp)^\perp = \mathcal{Y}$, we have the required result. $\square$

## 13.2. Dynamical system duality

*(This subsection is based on Section 12 and the section on dynamical systems in [19].)*

Dynamical system duality is based on the notion of adjoint dynamical system, which itself is based on vector space duality, since we have $\mathcal{Y}_{W\dot{W}UY'}^a := (\mathcal{Y}_{W\dot{W}UY})^\perp (-\dot{W})W(-Y')U'$.

**Problem 47.** The duals of the following 'primal' statements about dynamical systems are given in the succeeding paragraph by interchanging $W, \dot{W}$; $U, Y$; $\subseteq, \supseteq; +, \cap; \circ, \times; \leftrightarrow, \Leftrightarrow$. (Note that $(\mathcal{Y}_{SP} \Leftrightarrow \mathcal{Y}_P) = (\mathcal{Y}_{SP} \leftrightarrow \mathcal{Y}_P)$.) Assuming that the primal statements are true, prove the duals.

### Primal

1. $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$.

2. If $\mathcal{Y}_{WU} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU$, then $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W}$.

3. Let $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$.

If $\mathcal{Y}_{WU} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU$, then $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W}$.

4. Let $\mathcal{Y}_W$ be controlled invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)^\dot{W} \supseteq \mathcal{Y}_W$. Then it is conditioned invariant in $\mathcal{Y}_{W\dot{W}}$ iff $\mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_W)^\dot{W}$.

5. Let $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_W$ be such that $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}}$. Then $\mathcal{Y}_W$ is controlled invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U$, where $\mathcal{Y}_{WU} := ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}$.

### Dual

1. $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W$.

2. If $\mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y$ then $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}$.

3. Let $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W$.

If $\mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y$, then $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}$.
4. Let $\mathcal{Y}_W$ be conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$. Then it is controlled invariant in $\mathcal{Y}_{W\dot{W}}$ iff $\mathcal{Y}_{W\dot{W}} \circ W \supseteq \mathcal{Y}_W$.

5. Let $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{Y}_W$ be such that $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_W$. Then $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) \leftrightarrow \mathcal{Y}_Y$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_{\dot{W}Y}) \leftrightarrow \mathcal{Y}_Y$, where $\mathcal{Y}_{\dot{W}Y} := ((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_Y) \leftrightarrow \mathcal{Y}_W$.

Answer

We use the facts that $\mathcal{Y}_{W\dot{W}U'Y'} \coloneq (\mathcal{Y}_{W\dot{W}UY})_{(-\dot{W})W(-Y')U'}$, $\mathcal{Y}_{W\dot{W}}^a \coloneq (\mathcal{Y}_{W\dot{W}}^{\perp})_{(-\dot{W})W}$, $\mathcal{Y}_{WU'}^a \coloneq (\mathcal{Y}_{\dot{W}Y}^{\perp})_{WU'}$, $\mathcal{Y}_{\dot{W}Y'}^a \coloneq (\mathcal{Y}_{WU}^{\perp})_{(-\dot{W})(-Y')} = (\mathcal{Y}_{WU}^{\perp})_{\dot{W}Y'}$, and the facts that $(\mathcal{Y}^{\perp})^{\perp} = \mathcal{Y}$, $((\mathcal{Y}_{W\dot{W}UY}^{\perp})^a)^a = (\mathcal{Y}_{W\dot{W}U'Y'}^a)^a = \mathcal{Y}_{W\dot{W}UY}$.

1. The dual of $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$ is
$$ \mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \text{ and } \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W. $$

Reason: Step 1. The preliminary dual is obtained by replacing every space by its adjoint.

$$ \mathcal{Y}_{W\dot{W}U'Y'}^a \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}^a \text{ and } \mathcal{Y}_{W\dot{W}U'Y'}^a \circ W\dot{W}U' \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}}^a \times \dot{W}. $$

Step 2. Next, perform the adjoint operation interchanging variables using the rules $W$ to $-\dot{W}$, $\dot{W}$ to $W$, $U'$ to $-Y$, $Y'$ to $U$, $\circ$ to $\times$, $\times$ to $\circ$, $\subseteq$ to $\supseteq$, and $\supset$ to $\subseteq$,

$$ (\mathcal{Y}_{W\dot{W}U'Y'}^a \circ W\dot{W})^a \subseteq (\mathcal{Y}_{W\dot{W}}^a)^a \text{ and } (\mathcal{Y}_{W\dot{W}U'Y'}^a \circ W\dot{W}U' \times \dot{W})^a \supseteq (\mathcal{Y}_{W\dot{W}}^a \times \dot{W})^a, $$

i.e., $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W$.

2. The dual of $\mathcal{Y}_{WU} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU;"$ is

$$ \mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y. $$

Reason: Step 1. The preliminary dual is obtained by replacing every space by its adjoint.

$$ \mathcal{Y}_{WU'}^a = (\mathcal{Y}_{W\dot{W}U'Y'}^a \cap \mathcal{Y}_{W\dot{W}}^a) \circ WU'; $$

Step 2. Next, perform the adjoint operation interchanging variables using the rules $W$ to $-\dot{W}$, $\dot{W}$ to $W$, $U'$ to $-Y$, $Y'$ to $U$,

$$ (\mathcal{Y}_{WU'}^a)^a = ((\mathcal{Y}_{W\dot{W}U'Y'}^a \cap \mathcal{Y}_{W\dot{W}}^a) \circ WU')^a; $$

i.e.,
$$ \mathcal{Y}_{\dot{W}Y} = ((\mathcal{Y}_{W\dot{W}U'Y'}^a \cap \mathcal{Y}_{W\dot{W}}^a)^a \times WU')_{(-\dot{W})(-Y)} $$

$$ = ((\mathcal{Y}_{W\dot{W}U'Y'}^a)^a + (\mathcal{Y}_{W\dot{W}}^a)^a) \times WU'_{(-\dot{W})(-Y)} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y, $$

(noting that $\mathcal{Y}_{(-\dot{W})(-Y)} = \mathcal{Y}_{\dot{W}Y}$).

Thus we get $\mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y.$

The dual of $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W};$ is

$$ \mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}; $$

Reason: We use the two step process as before.

Step 1. $\mathcal{Y}_{W\dot{W}}^a = (\mathcal{Y}_{W\dot{W}UY}^a \cap \mathcal{Y}_{WU}^a) \circ W\dot{W};$

Step 2. $$(\mathcal{Y}_{W\dot{W}}^a)^a = \mathcal{Y}_{W\dot{W}} = ((\mathcal{Y}_{W\dot{W}UY}^a \cap \mathcal{Y}_{WU}^a) \circ W\dot{W})^a = (\mathcal{Y}_{W\dot{W}UY}^a \cap \mathcal{Y}_{WU}^a)^a \times W\dot{W}$$

$$ = ((\mathcal{Y}_{W\dot{W}UY}^a)^a + (\mathcal{Y}_{WU}^a)^a) \times W\dot{W} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}; $$

Thus we get $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}.$

3. The dual statement is

$$ \text{Let } \mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \text{ and } \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W. $$

If $\mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y$, then $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}.$

Reason: Suppose a set $\alpha$ of primal statements imply a primal statement A. Now the set $\alpha$ of primal statements are true iff the set $\beta$ of their duals are true and the primal statement A is true iff its dual B is true. Therefore the set $\beta$ implies the statement B.
In parts 1 and 2 above we have seen that the duals of a set $\alpha$ of primal statements are a set $\beta$ of dual statements and the dual of A is B. The sets $\alpha, \beta$ and statements A, B are as below.

$$ \alpha = \{ \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}; \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W} \}. $$

$$ A = \text{If } \mathcal{Y}_{WU} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU, \text{ then } \mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W}. $$

$$ \beta = \{ \mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}; \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W; \} $$

$$ B = \text{If } \mathcal{Y}_{\dot{W}Y} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y, \text{ then } \mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W}. $$

4. The dual statement is $C = $" Let $\mathcal{Y}_W$ be conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$. Then it is controlled invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$ iff $\mathcal{Y}_{\dot{W}W} \circ W \supseteq \mathcal{Y}_W$."

Reason: The following are dual pairs:

$$ \text{(primal)} \quad \mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W, \quad \text{(dual)} \quad \mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}. $$

$$ \text{(primal)} \quad \mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}, \quad \text{(dual)} \quad \mathcal{Y}_{W\dot{W}} \circ W \supseteq \mathcal{Y}_W. $$

The truth of the above dual statement C follows arguing as in part 3 above.

5. The dual statement is $D = $"Let $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{Y}_W$ be such that $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_W$. Then $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) \leftrightarrow \mathcal{Y}_Y$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_{\dot{W}Y}) \leftrightarrow \mathcal{Y}_Y$, where

$$ \mathcal{Y}_{\dot{W}Y} = ((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_Y) \leftrightarrow \mathcal{Y}_W. $$

Reason: The following are dual pairs:

$$ \text{(primal)} \quad \text{Let } \mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_W \text{ be such that } \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}}, $$

$$ \text{(dual)} \quad \text{Let } \mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_W \text{ be such that } \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_W. $$

$$ \text{(primal)} \quad \mathcal{Y}_W \text{ is controlled invariant in } (\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U; $$

$$ \text{(dual)} \quad \mathcal{Y}_W \text{ is conditioned invariant in } (\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) \leftrightarrow \mathcal{Y}_Y. $$

$$ \text{(primal)} \quad \mathcal{Y}_W \text{ is both controlled invariant and conditioned invariant in } ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U; $$

$$ \text{(dual)} \quad \mathcal{Y}_W \text{ is both controlled invariant and conditioned invariant in } ((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_{\dot{W}Y}) \leftrightarrow \mathcal{Y}_Y. $$

$$ \text{(primal)} \quad \mathcal{Y}_{WU} = ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}. $$

$$ \text{(dual)} \quad \mathcal{Y}_{\dot{W}Y} = ((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_Y) \leftrightarrow \mathcal{Y}_W. $$

The truth of the above dual statement D follows arguing as in part 3 above.

### 13.3. Capability of state feedback and output injection

*(This subsection is based on Section 12, Subsection 13.2.)*

**Problem 48.** Consider the following question.

'given a desired autonomous dynamical system $\mathcal{Y}_{W\dot{W}}$, can we obtain it from a specified dynamical system $\mathcal{Y}_{W\dot{W}UY}$ by state feedback and if so, how to obtain it?'

In the notation of this chapter, this question is solved as follows.

**Theorem 39.** Given $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{W\dot{W}}$, there exists $\mathcal{Y}_{WU}$ such that $(\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W} = \mathcal{Y}_{W\dot{W}}$ iff

$$ \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supseteq \mathcal{Y}_{W\dot{W}}; \quad \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}. \qquad (51) $$

Let the conditions in Equation 51 hold and let $\mathcal{Y}_{WU} := (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU$.

$$ \text{Then } (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W} = \mathcal{Y}_{W\dot{W}}. $$

Prove the result using Theorem 1.
Answer

Theorem 39 follows directly from Theorem 7. However, it can also be proved using Theorem 1. This proof is based on the following simple lemma which states that feedback can be reduced to the '$\leftrightarrow$' operation through a simple trick. This makes Theorem 1 applicable here. The proof of the lemma is omitted.

**Lemma 40.** Let $I_{AA'}$ denote the space of all $(f_A, f_{A'})$, where $A'$ is a copy of $A$ disjoint from $W \uplus \dot{W} \uplus U \uplus Y$. We then have

$$ (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W \dot{W} = [(\mathcal{Y}_{W\dot{W}UY} \cap I_{WW'}) \leftrightarrow ((\mathcal{Y}_{WU})_{W'U} \oplus \mathcal{F}_Y)]_{W\dot{W}}. $$

Note that $\mathcal{Y}_{W\dot{W}UY} \cap I_{WW'}$ has $(f_W, f_W, g_{\dot{W}UY})$ as a member iff $(f_W, g_{\dot{W}UY})$ is a member of $\mathcal{Y}_{W\dot{W}UY}$.

*Proof.* (of Theorem 39) Let us denote $\mathcal{Y}_{W\dot{W}UY} \cap I_{WW'}$ by $\mathcal{Y}_{W\dot{W}'UY}$, and $(\mathcal{Y}_{WU})_{W'U}$ by $\mathcal{Y}_{W'U}$. Thus, by Lemma 40, we need to examine the conditions under which there exists $\mathcal{Y}_{W'U}$ such that

$$ \mathcal{Y}_{WW'\dot{W}UY} \leftrightarrow (\mathcal{Y}_{W'U} \oplus \mathcal{F}_Y) = \mathcal{Y}_{W\dot{W}}, \text{ i.e., such that } \mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \leftrightarrow \mathcal{Y}_{W'U} = \mathcal{Y}_{W\dot{W}}. $$

By Theorem 1, this happens iff

$$ (a) \quad \mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \circ W\dot{W} \supset \mathcal{Y}_{W\dot{W}}, \text{ and } (b) \quad \mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}. $$

The condition (a) is one of the conditions in Equation 51, since $\mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \circ W\dot{W} = \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}$. We now examine (b).

$$ \begin{aligned} & \text{It is easy to see that } \mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \times W\dot{W} \\ &= \mathbf{0}_W \oplus \mathcal{Y}_{WW'\dot{W}UY} \circ W'\dot{W}U \times \dot{W} = \mathbf{0}_W \oplus \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W}. \end{aligned} $$

Therefore, $\mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}$ iff

$$ \mathbf{0}_W \oplus \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathbf{0}_W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}, \text{ i.e., iff} $$

$$ \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}, \text{ yielding the second condition in Equation 51.} $$

Next, again by Theorem 1, if such $\mathcal{Y}_{W'U}$ exists, i.e., if conditions in Equation 51 hold, a possible candidate for $\mathcal{Y}_{W'U}$ is $\mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \leftrightarrow \mathcal{Y}_{W\dot{W}}$

$$ = ((\mathcal{Y}_{WW'\dot{W}UY} \circ WW'\dot{W}U \cap \mathcal{Y}_{W\dot{W}}) \circ W'U $$

$$ = ((\mathcal{Y}_{WW'\dot{W}UY} \circ W'\dot{W}U) \cap (\mathcal{Y}_{W\dot{W}})_{W'\dot{W}}) \circ W'U $$

$$ = ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{W\dot{W}}) \circ WU)_{W'U} = ((\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU)_{W'U}, $$

$$ \text{i.e., a possible candidate for } \mathcal{Y}_{WU} \text{ is } (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU. \quad \square $$

**Problem 49.** Consider the following question.

'given a desired autonomous dynamical system $\mathcal{Y}_{W\dot{W}}$, can we obtain it from a specified dynamical system $\mathcal{Y}_{W\dot{W}'UY}$ by output injection and if so, how to obtain it?'

In the notation of this chapter, this question is solved as follows.

**Theorem 41.** Given $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{Y}_{W\dot{W}}$, there exists $\mathcal{Y}_{\dot{W}Y}$ such that $(\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W} = \mathcal{Y}_{W\dot{W}}$ iff

$$ \mathcal{Y}_{W\dot{W}UY} \times W\dot{W} \subseteq \mathcal{Y}_{W\dot{W}}; \quad \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W. \quad (52) $$

Let the conditions in Equation 52 hold and let $\mathcal{Y}_{\dot{W}Y} := (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{W\dot{W}}) \times \dot{W}Y$.

$$ \text{Then } (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W\dot{W} = \mathcal{Y}_{W\dot{W}}. $$

Prove the result.

Answer

Theorem 41 follows from Theorem 39 using adjoint system duality as in Problem 47.
**Problem 50.** Let the desired autonomous system $\mathcal{Y}_{W\dot{W}}$ be defined by the equation $\dot{w} = Kw$ and the dynamical system $\mathcal{Y}_{W\dot{W}UY}$ by $\dot{w} = Aw + Bu, y = Cw$.

1. Suppose that $\mathcal{Y}_{W\dot{W}}$ can be obtained from $\mathcal{Y}_{W\dot{W}UY}$ by $(W,U)$- feedback. Give an algorithm for building the state feedback (output injection) in the form $u = Fw$.

2. Suppose that $\mathcal{Y}_{W\dot{W}}$ can be obtained from $\mathcal{Y}_{W\dot{W}UY}$ by $(\dot{W},Y)$- injection. Give an algorithm for building the output injection in the form $\dot{w} = Gy$.

The algorithms should only use representative matrices for $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{Y}_{W\dot{W}}$ and not make explicit use of $A, B, C, K$.

Answer

1. For computing the state feedback matrix $F$, we first compute $\mathcal{Y}_{WU} := (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU$, as in Theorem 39. For $\mathcal{Y}_{WU}$, compute a representative matrix of the form $\begin{pmatrix} R_{1w} & R_{1u} \\ 0_{2w} & R_{2u} \end{pmatrix}$, where the rows of $R_{1w}, R_{2u}$ are linearly independent. Since $\mathcal{Y}_{W\dot{W}}$ is defined by $\dot{w} = Kw$, and $\mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W\dot{W}$, the rows $(0_{2w}|R_{2u})$ correspond to the situation where both $w = 0$ and $\dot{w} = 0$ and therefore can be dropped from $\mathcal{Y}_{WU}$ without affecting the LHS of the equation. So we may assume wlog that $\mathcal{Y}_{WU}$ has the representative matrix of the form $(R_{1w} \ R_{1u})$. Next, we have the necessary condition $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} \supset \mathcal{Y}_{W\dot{W}}$ for $\mathcal{Y}_{W\dot{W}}$ to be obtainable from $\mathcal{Y}_{W\dot{W}UY}$ by $(W,U)$- feedback. It follows that $\mathcal{Y}_{WU} \circ W = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{W\dot{W}}) \circ WU \circ W \supset \mathcal{Y}_{W\dot{W}} \circ W$, so that rows of $R_{1w}$ span $\mathcal{F}_W$, i.e., $\mathcal{Y}_{WU}$ has a representative matrix of the form $(I_w|F^T)$, where $F^T = R_{1w}^{-1}R_{1u}$.

2. For computing the output injection matrix $G$, we work with the adjoint systems $\mathcal{Y}_{W\dot{W}}^a := (\mathcal{Y}_{W\dot{W}})_{(-\dot{w})W}$ and $\mathcal{Y}_{W\dot{W}UY}^a := (\mathcal{Y}_{W\dot{W}UY})_{(-\dot{W})W(-Y')U'}$ compute the state feedback matrix $\hat{F}$, as in part (a). The required matrix $G = -\hat{F}^T$. We note that $\mathcal{Y}_{W\dot{W}}^a$ is defined by $\dot{w} = K^T w$ and $\mathcal{Y}_{W\dot{W}UY}^a$ is defined by $\dot{w} = A^T w + C^T u, y = B^T w$. As before, we only need the representative matrices for $\mathcal{Y}_{W\dot{W}}^a, \mathcal{Y}_{W\dot{W}UY}^a$ and do not need the matrices $A, B, C, K$, explicitly.

### 13.4. Controlled and conditioned invariant spaces strengthened through feedback and injection

*(This subsection is based on Section 12, Subsection 13.2.)*

**Problem 51.** Prove the following result which describes how to use $(W,U)$- feedback to (implicitly) make a controlled invariant space into a conditioned invariant space without losing its controlled invariance.

**Theorem 42.** Let $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_W$ be such that $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} U \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}} := (\mathcal{Y}_W)_{\dot{W}}$.

Then $\mathcal{Y}_W$ is controlled invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} U) \leftrightarrow \mathcal{Y}_U$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} U) \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U$, where

$$ \mathcal{Y}_{WU} := ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W} U) \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}. $$

Answer

1. We need the following result, which is a consequence of a mild variation of Theorem 1, and is from [19].

**Theorem 43.** 1. Let $\mathcal{Y}_W$ be conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$. Then it is controlled invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$ iff $\mathcal{Y}_{\dot{W}W} \circ W \supseteq \mathcal{Y}_W$.

2. Let $\mathcal{Y}_W$ be controlled invariant in $\mathcal{Y}_{W\dot{W}}$. Then it is conditioned invariant in $\mathcal{Y}_{W\dot{W}}$ iff $\mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$.
*Proof.* (Proof of Theorem 42) Let $\mathcal{Y}_{WU} := ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}$. We claim that $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U$.

(i) **Claim:** $\mathcal{Y}_W$ is controlled invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U$.

We know that $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}} \supseteq \mathcal{Y}_W$.

Suppose $w \in \mathcal{Y}_W$. Then there exists $(w, \dot{w}, u) \in (\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U)$ with $\dot{w} \in \mathcal{Y}_{\dot{W}}$, $u \in \mathcal{Y}_U$.

But then by the definition of $\mathcal{Y}_{WU}$, we have $(w, u) \in \mathcal{Y}_{WU}$.

Hence $w \in ((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}$,

i.e., $\mathcal{Y}_W$ is controlled invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU})) \leftrightarrow \mathcal{Y}_U)$.

(ii) **Claim:** $\mathcal{Y}_W$ is conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U)$.

By Theorem 43, we need to show that $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U) \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}}$.

Let $\dot{w} \in \text{LHS. Then there exists } (0, \dot{w}, u) \in \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}$ such that $u \in \mathcal{Y}_U$.

Now $(0, u) \in \mathcal{Y}_{WU} := (\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}$.

Hence there exists $\dot{w}_1 \in \mathcal{Y}_{\dot{W}}$ such that $(0, \dot{w}_1, u) \in \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}$.

It follows that $(0, \dot{w} - \dot{w}_1, 0) \in \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU} \subseteq \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U$.

Hence $\dot{w} - \dot{w}_1 \in \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}}$. We conclude that $\dot{w} \in \mathcal{Y}_{\dot{W}}$ since $\dot{w}_1 \in \mathcal{Y}_{\dot{W}}$.

On the other hand, if $\mathcal{Y}_W$ is controlled invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U)$,

i.e., $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}} \supseteq \mathcal{Y}_W$,

it is immediate that $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}} \supseteq \mathcal{Y}_W$,

i.e., that $\mathcal{Y}_W$ is controlled invariant in $((\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U) \leftrightarrow \mathcal{Y}_U)$. $\square$

**Problem 52.** Prove the following result which describes how to use $(\dot{W}, Y)$- injection to (implicitly) make a conditioned invariant space into a controlled invariant space without losing its controlled invariance.

**Theorem 44.** Let $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{Y}_W$ be such that $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_W$. Then $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) \leftrightarrow \mathcal{Y}_Y$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_{\dot{W}}) \leftrightarrow \mathcal{Y}_Y$, where $\mathcal{Y}_{\dot{W}} := ((\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y) + \mathcal{Y}_Y) \leftrightarrow \mathcal{Y}_W$.

*Proof.* The theorem follows from Theorem 42 using adjoint system duality as in Problem 47. $\square$

**Problem 53.** Let the dynamical system $\mathcal{Y}_{W\dot{W}U}$ be defined through $\dot{w} = Aw + Bu$. Show that $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}U} \leftrightarrow \mathcal{Y}_U$, i.e., $\mathcal{Y}_{W\dot{W}U} \leftrightarrow \mathcal{Y}_U \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$, where $\mathcal{Y}_U = \mathcal{F}_U$, iff there exists a matrix $F$, such that $(A + BF)\mathcal{Y}_W \subseteq \mathcal{Y}_W$.

*Answer*

Let $\mathcal{Y}_{W\dot{W}U}$ be defined by $\dot{w} = Aw + Bu$. If there exists a matrix such that $(A + BF)\mathcal{Y}_W \subseteq \mathcal{Y}_W$, it is clear that $A\mathcal{Y}_W \subseteq \mathcal{Y}_W + \mathcal{B}$, where $\mathcal{B}$ is the span of the columns of $B$. We have seen in Problem 43, that this latter condition is equivalent to controlled invariance of $\mathcal{Y}_W$ in $\mathcal{Y}_{W\dot{W}U} \leftrightarrow \mathcal{F}_U$.

Thus if $\mathcal{Y}_{W\dot{W}U}$ is defined by $\dot{w} = Aw + Bu$, then $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}U} \circ W\dot{W}$.

We now examine the converse. We have $\mathcal{Y}_{W\dot{W}U} \times \dot{W} = \mathbf{0}_{\dot{W}}$, when the system is defined by $\dot{w} = Aw + Bu$, so that $\mathcal{Y}_{W\dot{W}U} \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}}$, for any subspace $\mathcal{Y}_{\dot{W}}$. Therefore, we have by Theorem 42, that the space $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}U} \leftrightarrow \mathcal{F}_U$ iff $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $\mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}_{WU} \leftrightarrow \mathcal{F}_U$, where $\mathcal{Y}_{WU} := (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{F}_U) \leftrightarrow (\mathcal{Y}_W)_{\dot{W}}$.
We will now show, we may take without loss of generality, that $\mathcal{Y}_{WU} \times U = \mathbf{0}_U$, which means $\mathcal{Y}_{WU}$ can be defined by $u = F^{small}w, w \in \mathcal{Y}_{WU} \circ W$.

Suppose $\mathcal{Y}_{WU} \subseteq \mathcal{Y}'_{WU}$. It is then clear that $\mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}_{WU} \leftrightarrow \mathcal{F}_U \subseteq \mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}'_{WU} \leftrightarrow \mathcal{F}_U$, so that $\mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}_{WU} \leftrightarrow \mathcal{F}_U \leftrightarrow \mathcal{Y}_W \subseteq \mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}'_{WU} \leftrightarrow \mathcal{F}_U$, i.e., if $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}'_{WU} \leftrightarrow \mathcal{F}_U)$, it is also conditioned invariant in $(\mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}_{WU} \leftrightarrow \mathcal{F}_U)$ and $\mathcal{Y}_{WU} \times U = \mathbf{0}_U$, which means $\mathcal{Y}_{WU}$ can be defined by $u = F^{small}w, w \in \mathcal{Y}_{WU} \circ W$. Let $\hat{\mathcal{Y}}_{WU}$ be defined by $u = F^{small}w, w \in \mathcal{Y}_{WU} \circ W$, $u = F''w, w \in \mathcal{Y}'_W$, where $\mathcal{Y}'_W$ is such that $\mathcal{Y}_{WU} \circ W \cap \mathcal{Y}'_W = \mathbf{0}_W$, $\mathcal{Y}_{WU} \circ W + \mathcal{Y}'_W = \mathcal{F}_W$. It is clear that $\hat{\mathcal{Y}}_{WU}$ is defined by $u = Fw$, for some $F$ and $\mathcal{Y}_{W\dot{W}U} \cap \hat{\mathcal{Y}}_{WU} \leftrightarrow \mathcal{F}_U \leftrightarrow \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}U} \cap \mathcal{Y}_{WU} \leftrightarrow \mathcal{F}_U \leftrightarrow \mathcal{Y}_W$. Therefore $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}U} \cap \hat{\mathcal{Y}}_{WU} \leftrightarrow \mathcal{F}_U$.

Now $\mathcal{Y}_{W\dot{W}U}$ is defined by $\dot{w} = Aw + Bu$ and $\hat{\mathcal{Y}}_{WU}$ is defined by $u = Fw$. Therefore, in $\mathcal{Y}_{W\dot{W}U} \cap \hat{\mathcal{Y}}_{WU} \leftrightarrow \mathcal{F}_U$, when $w \in \mathcal{Y}_W$ we must have $\dot{w} = Aw + Bu = Aw + BFw = (A + BF)w$, so that we have $((A + BF)(\mathcal{Y}_W))_{\dot{W}} = \mathcal{Y}_{W\dot{W}U} \cap \hat{\mathcal{Y}}_{WU} \leftrightarrow \mathcal{F}_U \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$, i.e., $(A + BF)(\mathcal{Y}_W) \subseteq \mathcal{Y}_W$.

**Problem 54.** Let the dynamical system $\mathcal{Y}_{W\dot{W}Y}$ be defined through $\dot{w} = Aw, y = Cw$ i.e., $M = Y, U = \emptyset$. Show that $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}Y} \leftrightarrow \mathcal{Y}_Y$, i.e., $\mathcal{Y}_{W\dot{W}Y} \leftrightarrow \mathcal{Y}_Y \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$, where $\mathcal{Y}_Y = \mathbf{0}_Y$, iff there exists a matrix $G$, such that $(A - GC)\mathcal{Y}_W \subseteq \mathcal{Y}_W$.

Answer

This problem is dual to Problem 53 above and can be solved by working with the adjoint as in Problem 47. The following are dual pairs:

1 (primal) $\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} U \times \dot{W} \subseteq \mathcal{Y}_{\dot{W}} :\equiv (\mathcal{Y}_W)_{\dot{W}}$;
(dual) $\mathcal{Y}_{W\dot{W}UY} \times W \dot{W} Y \circ W \supseteq \mathcal{Y}_W$.

2 (primal) $\mathcal{Y}_W$ is controlled invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} U) \leftrightarrow \mathcal{Y}_U$;
(dual) $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \times W \dot{W} Y) \leftrightarrow \mathcal{Y}_Y$.

3 (primal) $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $(\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} U \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{Y}_U$ where $\mathcal{Y}_{WU} :\equiv ((\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} U) \cap \mathcal{Y}_U) \leftrightarrow \mathcal{Y}_{\dot{W}}$;
(dual) $\mathcal{Y}_W$ is both controlled invariant and conditioned invariant in $((\mathcal{Y}_{W\dot{W}UY} \times W \dot{W} Y) + \mathcal{Y}_{\dot{W}Y}) \leftrightarrow \mathcal{Y}_Y$, where $\mathcal{Y}_{\dot{W}Y} :\equiv ((\mathcal{Y}_{W\dot{W}UY} \times W \dot{W} Y) + \mathcal{Y}_Y) \leftrightarrow \mathcal{Y}_W$.

4 (primal) $\mathcal{Y}_{WU}$ is defined by $u = Fw$;
(dual) $\mathcal{Y}_{\dot{W}Y}$ is defined by $\dot{w} = -F^T y$.

5 (primal) $\mathcal{Y}_U = \mathcal{F}_U$;
(dual) $\mathcal{Y}_Y = \mathbf{0}_Y$.

6 (primal) $\dot{w} = Aw + Bu$;
(dual) $\dot{w} = A^T w, y = B^T w$.

7 (primal) $(A + BF)\mathcal{Y}_W \subseteq \mathcal{Y}_W$;
(dual) $(A^T - F^T B^T)\mathcal{Y}_W \subseteq \mathcal{Y}_W$.

A statement is true iff the dual statement is true. A set of statements imply a statement $D$ iff the dual set of statements imply the dual statement $D^d$.

Thus the statement "in the dynamical system $\mathcal{Y}_{W\dot{W}U}$ defined through $\dot{w} = Aw + Bu$, $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}U} \leftrightarrow \mathcal{Y}_U$, where $\mathcal{Y}_U = \mathcal{F}_U$, iff there exists a matrix $F$, such that $(A + BF)\mathcal{Y}_W \subseteq \mathcal{Y}_W$;" is true iff the statement " in the dynamical system $\mathcal{Y}_{W\dot{W}Y}$ defined through $\dot{w} = A^T w, y = B^T w$, $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}Y} \leftrightarrow \mathcal{Y}_Y$, where $\mathcal{Y}_Y = \mathbf{0}_Y$, iff there exists a matrix $F^T$, such that $(A - F^T B^T)\mathcal{Y}_W \subseteq \mathcal{Y}_W$."
### 13.5. Self bounded controlled invariance and self hidden conditioned invariance

*(This subsection is based on Section 12, Subsection 13.2.)*

**Problem 55.** For a system defined in the state-input-output form $\dot{w} = Aw+Bu, y = Cw$, we say $\mathcal{Y}_W \subseteq \mathcal{E}_W$ is self bounded controlled invariant with respect to $\mathcal{E}_W$ ([1]) iff (a) $\mathcal{Y}_W$ is controlled invariant in the system; (b) if $\mathcal{Y}'_W \subseteq \mathcal{E}_W$ is any other controlled invariant space in the system, then $\mathcal{Y}'_W \cap \mathcal{B} \subseteq \mathcal{Y}_W$, where $\mathcal{B} := \text{Im}(B)$. What is a self bounded controlled invariant subspace in the notation of the present chapter?

**Answer**

Let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{F}_{UY} = \mathcal{Y}_{W\dot{W}UY} \circ W \dot{W}$. We note that $\mathcal{B} = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W$. Therefore, $\mathcal{Y}_W \subseteq \mathcal{E}_W$ is self bounded controlled invariant with respect to $\mathcal{E}_W$ iff (a) $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$; and (b) whenever $\mathcal{Y}'_W \subseteq \mathcal{E}_W$ is such that $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}'_W)_{\dot{W}} \supseteq \mathcal{Y}'_W$, we have $\mathcal{Y}'_W \cap (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \subseteq \mathcal{Y}_W$. ((b) is equivalent to the condition (b') the maximal controlled invariant space $\mathcal{Y}^{controlled}(\mathcal{E})$ of $\mathcal{Y}_{W\dot{W}}$ contained in $\mathcal{E}_W$ satisfies $\mathcal{Y}^{controlled}(\mathcal{E}) \cap (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \subseteq \mathcal{Y}_W$.)

**Problem 56.** For a system defined in the state-input-output form $\dot{w} = Aw+Bu, y = Cw$, we say $\mathcal{Y}_W \supseteq \mathcal{D}_W$ is self hidden conditioned invariant with respect to $\mathcal{D}_W$ ([1]) iff (a) $\mathcal{Y}_W$ is conditioned invariant in the system; (b) if $\mathcal{Y}'_W \supseteq \mathcal{D}_W$ is any other conditioned invariant space in the system, then $\mathcal{Y}'_W + \mathcal{C} \supseteq \mathcal{Y}_W$, where $\mathcal{C} := \text{Ker}(C)$.

What is a self hidden conditioned invariant subspace in the notation of the present chapter? Show that the notions of self bounded controlled invariant subspace and self hidden conditioned invariant subspace are duals of each other, with duality as in Problem 47.

**Answer**

Let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathbf{0}_{UY} = \mathcal{Y}_{W\dot{W}UY} \times W \dot{W}$. We note that $\mathcal{C} = (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}}$. Therefore, $\mathcal{Y}_W \supseteq \mathcal{D}_W$ is self hidden conditioned invariant with respect to $\mathcal{D}_W$ iff (a) $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$; and (b) whenever $\mathcal{Y}'_W \supseteq \mathcal{D}_W$ is such that $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}'_W \subseteq (\mathcal{Y}'_W)_{\dot{W}}$ we have $\mathcal{Y}'_W + (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_W$. ((b) is equivalent to the condition (b') the minimal conditioned invariant space $\mathcal{Y}^{conditioned}(\mathcal{D})$ of $\mathcal{Y}_{W\dot{W}}$ containing $\mathcal{D}_W$ satisfies $\mathcal{Y}^{conditioned}(\mathcal{D}) + \mathcal{Y}_{W\dot{W}} \circ W \supseteq \mathcal{Y}_W$.)

According to the notion of duality as in Problem 47, the following are dual pairs:

$$ \begin{aligned} \mathcal{Y}_{W\dot{W}UY} \circ W \dot{W}, \mathcal{Y}_{W\dot{W}UY} \times W \dot{W}; &\quad \mathcal{B} = (\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} \times \dot{W})_W, \mathcal{C} = \mathcal{Y}_{W\dot{W}UY} \times W \dot{W} \circ W; \\ \mathcal{Y}_W \subseteq \mathcal{E}_W, \mathcal{Y}_W \supseteq \mathcal{D}_W &\quad (\text{taking } \mathcal{E}_W = \mathcal{D}_W, \mathcal{Y}'_W \subseteq \mathcal{E}_W \text{ to be equivalent to } (\mathcal{Y}'_W)_{\dot{W}} \supseteq (\mathcal{E}_W)_{\dot{W}}); \end{aligned} $$

$$ \begin{aligned} \mathcal{Y}'_W \cap (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \subseteq \mathcal{Y}_W, \mathcal{Y}'_W + (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_W &\quad (\text{taking } (\mathcal{Y}'_W)_{\dot{W}} + (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq (\mathcal{Y}_W)_{\dot{W}} \text{ to be equivalent to } \mathcal{Y}'_W + (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_W); \end{aligned} $$

If one makes the substitutions as above in the statements of the answer to Problem 55, it is clear that we get the statements of the answer to Problem 56.

### 13.6. Stabilizability and detectability

*(This subsection is based on Section 12, Subsection 13.2.)*

In order to talk of stability, we need to bring in eigenvalues of the matrix $A$ of the state variable formulation of the dynamical system. In the present paper, thus far, we have not developed the necessary formalism to speak of polynomials associated with $(\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{UY})$. (This is done in Section 15.) As a temporary measure we can speak in terms of eigen spaces corresponding to eigenvalues with negative real parts and nonnegative real parts and take these to be specified.

**Problem 57.** We say a space $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$ iff it is both controlled and conditioned invariant in $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}$. Let $\mathcal{Y}_W^1, \mathcal{Y}_W^2$, be two invariant spaces of $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$. We say that $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ is a **complementary pair** iff $\mathcal{Y}_W^1 \cap \mathcal{Y}_W^2 = \mathbf{0}_W$ and $\mathcal{Y}_W^1 + \mathcal{Y}_W^2 = \mathcal{F}_W$. For the dynamical system defined by $\dot{w} = Aw + Bu, y = Cw, \mathcal{Y}_W^1$ could be the eigenspace of $A$ corresponding
to all its eigenvalues with negative real parts and $\mathcal{Y}_W^2$ could be the eigenspace of A corresponding to all its eigenvalues with nonnegative real parts.

We say a controlled invariant space $\mathcal{Y}_W$, of $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$, is **internally stable** relative to the complementary pair $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ iff $\mathcal{Y}_W \subseteq \mathcal{Y}_W^1$. For the dynamical system defined by $\dot{w} = Aw + Bu, y = Cw$, taking $\mathcal{Y}_{UY} := \mathbf{0}_U \oplus \mathcal{F}_Y$, if $\mathcal{Y}_W^1$ is the eigenspace of A corresponding to all its eigenvalues with negative real parts, this means $A|_{\mathcal{Y}_W}$ has all its eigenvalues with negative real parts.

We say a conditioned invariant space $\mathcal{Y}_W$, of $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$, is **externally stable** relative to the complementary pair $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ iff $\mathcal{Y}_W \supseteq \mathcal{Y}_W^2$. For the above dynamical system this means the map $A/\mathcal{Y}_W := A|_{\mathcal{F}_W/\mathcal{Y}_W}$ induced by A on the quotient space $\mathcal{F}_W/\mathcal{Y}_W$ has all its eigenvalues with negative real parts.

Using this notation define 'stabilizable' and 'detectable' in our formalism and show that they are dual notions.

### Answer

We note that the following are (adjoint) dual pairs:

(primal) $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)$,

(dual) $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)$;

(primal) minimal conditioned invariant space containing $(\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathbf{0}_W \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y))_W$,

(dual) maximal controlled invariant space contained in $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{F}_{\dot{W}} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)$.

Further (see subsection on adjoint systems in [19])

* $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ is a complementary pair of invariant spaces of $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}$ iff $((\mathcal{Y}_W^2)^\perp, (\mathcal{Y}_W^1)^\perp)$ is a complementary pair of invariant spaces of $\mathcal{Y}_{W\dot{W}U'Y'}^a \leftrightarrow (\mathcal{Y}_{U'Y'}^\perp)_{(-Y')U'}$.

* $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}$ and is contained in $\hat{\mathcal{Y}}_W$ iff $\mathcal{Y}_W^\perp$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}U'Y'}^a \leftrightarrow (\mathcal{Y}_{U'Y'}^\perp)_{(-Y')U'}$ and contains $\hat{\mathcal{Y}}_W^\perp$,

* $\mathcal{Y}_W$ is controlled (conditioned) invariant in $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}$ and is internally (externally) stable relative to the complementary pair $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ iff $\mathcal{Y}_W^\perp$ is conditioned (controlled) invariant in $\mathcal{Y}_{W\dot{W}U'Y'}^a \leftrightarrow (\mathcal{Y}_{U'Y'}^\perp)_{(-Y')U'}$ and is externally (internally) stable relative to the complementary pair $((\mathcal{Y}_W^2)^\perp, (\mathcal{Y}_W^1)^\perp)$.

It follows that the statements

‘controlled invariant space $\mathcal{Y}_W$, of $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$ is **internally stable** relative to the complementary pair $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$,’

‘conditioned invariant space $\mathcal{Y}_W$, of $\mathcal{Y}_{W\dot{W}UY}$ with respect to $\mathcal{Y}_{UY}$ is **externally stable** relative to the complementary pair $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$.’

are dual pairs.

We say $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)$ is **stabilizable** relative to $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ iff the minimal conditioned invariant space containing $(\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathbf{0}_W \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y))_W = (\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y) \leftrightarrow \mathbf{0}_W)_W$ is externally stable relative to $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$.

We say $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}$ is **detectable** relative to $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$ iff the maximal controlled invariant space contained in $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{F}_{\dot{W}} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y) = \mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y) \leftrightarrow \mathcal{F}_{\dot{W}}$ is internally stable relative to $(\mathcal{Y}_W^1, \mathcal{Y}_W^2)$.

It follows from the above discussion that stabilizability and detectability are dual notions.

We note that when the dynamical system $\mathcal{Y}_{W\dot{W}UY}$ is defined by $\dot{w} = Aw + Bu, y = Cw$, $(\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)) \circ W = \mathcal{F}_W \supseteq \mathcal{Y}_W$ for any conditioned invariant space $\mathcal{Y}_W$, so that by Theorem 43, $\mathcal{Y}_W$ is also controlled invariant. Next, $(\mathcal{Y}_{W\dot{W}UY} \leftrightarrow (\mathbf{0}_U \oplus \mathcal{F}_Y)) \times \dot{W} = \mathbf{0}_W \dot{\times} (\mathcal{Y}_W) \dot{\times} \dot{W}$ for any controlled invariant space $\mathcal{Y}_W$, so that by Theorem 43, $\mathcal{Y}_W$ is also conditioned invariant. Therefore, for the dynamical system defined by $\dot{w} = Aw + Bu, y = Cw$, in the above definitions of stabilizability and detectability, ‘controlled/conditioned invariance’ can be replaced by ‘invariance’.
### 13.7. Emulators

\*(This subsection is based on Section 12, Subsection 13.2.)

A dynamical system $\mathcal{Y}_{P\dot{P}UY} (\mathcal{Y}_{P\dot{P}})$ is said to be an **emulator of** $\mathcal{Y}_{W\dot{W}UY}, (\mathcal{Y}_{W\dot{W}},)$ $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$, iff $\mathcal{Y}_{W\dot{W}UY} = \mathcal{Y}_{P\dot{P}UY} \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) (\mathcal{Y}_{W\dot{W}} = \mathcal{Y}_{P\dot{P}} \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2))$ and $\mathcal{Y}_{WP}^1 \supseteq (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{WP}$. If both $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{P\dot{P}UY} (\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}})$ are emulators of each other $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$. We refer to them as **mutual emulators** $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$. We refer to them as **mutual emulators** $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$. We refer to them as **mutual emulators** $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$, as an $\mathcal{E}$-linkage for the mutual emulators $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{P\dot{P}UY} (\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}})$. If $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{P\dot{P}UY}$ are $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$, it is clear (using Theorem 3) that so are $\mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{Y}_{UY}, \mathcal{Y}_{P\dot{P}UY} \leftrightarrow \mathcal{Y}_{UY}$.

**Problem 58.** Prove the following result which shows that emulators of emulators continue to be emulators of the original dynamical system.

**Theorem 45.** Let $\mathcal{Y}_{W\dot{W}UY} (\mathcal{Y}_{W\dot{W}UY})$ be an emulator for $\mathcal{Y}_{P\dot{P}UY}, (\mathcal{Y}_{P\dot{P}UY})$ $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$ and let $\mathcal{Y}_{Q\dot{Q}UY} (\mathcal{Y}_{Q\dot{Q}})$ be an emulator for $\mathcal{Y}_{W\dot{W}UY}, (\mathcal{Y}_{W\dot{W}})$ $\mathcal{E}$-linked through $\mathcal{Y}_{QW}^1, \mathcal{Y}_{Q\dot{W}}^2$. Then $\mathcal{Y}_{Q\dot{Q}UY} (\mathcal{Y}_{Q\dot{Q}})$ is an emulator for $\mathcal{Y}_{P\dot{P}UY}, (\mathcal{Y}_{P\dot{P}})$ $\mathcal{E}$-linked through $(\mathcal{Y}_{QW}^1 \leftrightarrow \mathcal{Y}_{WP}^1), (\mathcal{Y}_{Q\dot{W}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2)$.

**Answer**

*Proof.* We will prove only the case for $\mathcal{Y}_{Q\dot{Q}UY}, \mathcal{Y}_{P\dot{P}UY}$, the case for $\mathcal{Y}_{Q\dot{Q}}, \mathcal{Y}_{P\dot{P}}$, having an identical proof. By the definition of an $\mathcal{E}$-linkage we have that

$$ \begin{aligned} \mathcal{Y}_{Q\dot{Q}UY} &= (\mathcal{Y}_{QW}^1 \oplus \mathcal{Y}_{\dot{Q}\dot{W}}^2) \leftrightarrow \mathcal{Y}_{W\dot{W}UY} \\ &= (\mathcal{Y}_{QW}^1 \oplus \mathcal{Y}_{\dot{Q}\dot{W}}^2) \leftrightarrow ((\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{Y}_{P\dot{P}UY}). \end{aligned} $$

By Theorem 3 (since no index set appears more than twice) we then have

$$ \mathcal{Y}_{Q\dot{Q}UY} = ((\mathcal{Y}_{QW}^1 \leftrightarrow \mathcal{Y}_{WP}^1) \oplus (\mathcal{Y}_{\dot{Q}\dot{W}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2)) \leftrightarrow \mathcal{Y}_{P\dot{P}UY}. $$

Since $(\mathcal{Y}_{WP}^1)_{\dot{W}\dot{P}} \supseteq \mathcal{Y}_{\dot{W}\dot{P}}^2, (\mathcal{Y}_{QW}^1)_{\dot{Q}\dot{W}} \supseteq \mathcal{Y}_{\dot{Q}\dot{W}}^2$, it is clear that $(\mathcal{Y}_{QW}^1 \leftrightarrow \mathcal{Y}_{WP}^1)_{\dot{Q}\dot{P}} \supseteq (\mathcal{Y}_{\dot{Q}\dot{W}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2)$. Therefore $\mathcal{Y}_{Q\dot{Q}UY}$ is an emulator for $\mathcal{Y}_{P\dot{P}UY}, \mathcal{E}$-linked through $(\mathcal{Y}_{QW}^1 \leftrightarrow \mathcal{Y}_{WP}^1) \oplus (\mathcal{Y}_{\dot{Q}\dot{W}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2)$. $\square$

**Problem 59.** Prove the following result which shows how state feedback and output injection operations transfer across mutual emulators.

**Theorem 46.** Let $\mathcal{Y}_{P\dot{P}UY}, \mathcal{Y}_{W\dot{W}UY}$ be mutual emulators, $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{WP}^2$.

Let $\mathcal{Y}_{WU}, \mathcal{Y}_{\dot{W}Y}, \mathcal{Y}_{PU}, \mathcal{Y}_{\dot{P}Y}$ be vector spaces on $W \oplus U, \dot{W} \oplus Y, P \oplus U, \dot{P} \oplus Y$ respectively such that $\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{WU} = \mathcal{Y}_{PU}, \mathcal{Y}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}Y} = \mathcal{Y}_{\dot{P}Y}$.

We then have the following.

$$ 1. (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow ((\mathcal{Y}_{P\dot{P}UY} \cap \mathcal{Y}_{PU}) \circ P \dot{P}) = (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W \dot{W}; $$

$$ 2. (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow ((\mathcal{Y}_{P\dot{P}UY} + \mathcal{Y}_{\dot{P}Y}) \times P \dot{P}) = (\mathcal{Y}_{W\dot{W}UY} + \mathcal{Y}_{\dot{W}Y}) \times W \dot{W}. $$

**Answer**

We need the following lemma and the succeeding result for the proof of this theorem.

**Lemma 47.** Let $\mathcal{Y}_{AB}, \mathcal{Y}_{BC}^1, \mathcal{Y}_{BC}^2$ be vector spaces with sets A, B, C, being pairwise disjoint. We then have the following.

$$ 1. \mathcal{Y}_{AB} \leftrightarrow (\mathcal{Y}_{BC}^1 \cap \mathcal{Y}_{BC}^2) = (\mathcal{Y}_{AB} \leftrightarrow \mathcal{Y}_{BC}^1) \cap (\mathcal{Y}_{AB} \leftrightarrow \mathcal{Y}_{BC}^2) \text{ if } \mathcal{Y}_{AB} \times B \subseteq \mathcal{Y}_{BC}^1 \times B. $$
2. $\mathcal{Y}_{AB} \leftrightarrow (\mathcal{Y}_{BC}^1 + \mathcal{Y}_{BC}^2) = (\mathcal{Y}_{AB} \leftrightarrow \mathcal{Y}_{BC}^1) + (\mathcal{Y}_{AB} \leftrightarrow \mathcal{Y}_{BC}^2)$ if $\mathcal{Y}_{AB} \circ B \supseteq \mathcal{Y}_{BC}^1 \circ B$.

See Problem 2, for the proof. The following result, which is a consequence of Theorem 1, is from [19].

**Theorem 48.** Let $\mathcal{Y}_{P\dot{P}UY}$ be an emulator of $\mathcal{Y}_{W\dot{W}UY}$, $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. We then have the following.

1. $\mathcal{Y}_{WP}^1 \circ W \supseteq \mathcal{Y}_{W\dot{W}UY} \circ W, \mathcal{Y}_{WP}^1 \times W \subseteq \mathcal{Y}_{W\dot{W}UY} \times W,$
$\mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W} \supseteq \mathcal{Y}_{W\dot{W}UY} \circ \dot{W}, \mathcal{Y}_{\dot{W}\dot{P}}^2 \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}UY} \times \dot{W}.$

2. $\mathcal{Y}_{W\dot{W}UY}$ is an emulator of $\mathcal{Y}_{P\dot{P}UY}$, $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. iff
$\mathcal{Y}_{WP}^1 \circ P \supseteq \mathcal{Y}_{P\dot{P}UY} \circ P, \mathcal{Y}_{WP}^1 \times P \subseteq \mathcal{Y}_{P\dot{P}UY} \times P,$
$\mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{P} \supseteq \mathcal{Y}_{P\dot{P}UY} \circ \dot{P}, \mathcal{Y}_{\dot{W}\dot{P}}^2 \times \dot{P} \subseteq \mathcal{Y}_{P\dot{P}UY} \times \dot{P}.$

*Proof.* (of Theorem 46)

1. Since $\mathcal{Y}_{W\dot{W}UY}, \mathcal{Y}_{P\dot{P}UY}$, are mutual emulators $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$, we have by Theorem 48,
$\mathcal{Y}_{WP}^1 \times W \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2 \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}UY} \times W \oplus \mathcal{Y}_{W\dot{W}UY} \times \dot{W};$
$\mathcal{Y}_{WP}^1 \times P \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2 \times \dot{P} \subseteq \mathcal{Y}_{W\dot{W}UY} \times P \oplus \mathcal{Y}_{W\dot{W}UY} \times \dot{P}.$

Now

$$ \begin{align*} & ((\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow (\mathcal{Y}_{P\dot{P}UY} \cap \mathcal{Y}_{PU})) \circ P \dot{P} = ((\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow ((\mathcal{Y}_{P\dot{P}UY} \cap (\mathcal{Y}_{PU} \oplus \mathcal{F}_{\dot{P}} \oplus \mathcal{F}_Y)) \circ P \dot{P} \\ &= ((\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{Y}_{P\dot{P}UY}) \cap ((\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow (\mathcal{Y}_{PU} \oplus \mathcal{F}_{\dot{P}} \oplus \mathcal{F}_Y)) \leftrightarrow \mathcal{F}_{UY}, \text{(using Lemma 47)} \\ &= (\mathcal{Y}_{W\dot{W}UY} \cap ((\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{PU}) \oplus (\mathcal{Y}_{\dot{W}\dot{P}}^2 \leftrightarrow (\mathcal{F}_{\dot{P}} \oplus \mathcal{F}_Y)) \leftrightarrow \mathcal{F}_{UY} \\ &= (\mathcal{Y}_{W\dot{W}UY} \cap (\mathcal{Y}_{WU} \oplus (\mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W}) \oplus \mathcal{F}_Y)) \leftrightarrow \mathcal{F}_{UY} \\ &= (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \leftrightarrow \mathcal{F}_{UY} \text{ (since } \mathcal{Y}_{W\dot{W}UY} \circ \dot{W}Y \subseteq (\mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W}) \oplus \mathcal{F}_Y \text{ by Theorem 48}), \text{ i.e.,} \\ &= (\mathcal{Y}_{W\dot{W}UY} \cap \mathcal{Y}_{WU}) \circ W \dot{W}. \end{align*} $$

2. The proof is dual to that of part 1 using part 2 of Lemma 47.

For notational convenience, let us denote $\mathcal{Y}_{W\dot{W}U'Y'}^a$ by $\hat{\mathcal{Y}}_{W\dot{W}U'Y'}$, $\mathcal{Y}_{P\dot{P}U'Y'}^a$ by $\hat{\mathcal{Y}}_{P\dot{P}U'Y'}$, $((\mathcal{Y}_{WP}^1)^\perp)_{\dot{W}\dot{P}}$ by $\hat{\mathcal{Y}}_{\dot{W}\dot{P}}^2$, $((\mathcal{Y}_{\dot{W}\dot{P}}^2)^\perp)_{WP}$ by $\hat{\mathcal{Y}}_{WP}^1$, $(\mathcal{Y}_{\dot{W}\dot{Y}}^\perp)_{WU'}$ by $\hat{\mathcal{Y}}_{WU'}$, $(\mathcal{Y}_{\dot{W}\dot{U}}^\perp)_{\dot{W}\dot{P}}$ by $\hat{\mathcal{Y}}_{\dot{W}\dot{Y}}$, $(\mathcal{Y}_{\dot{P}\dot{U}}^\perp)_{PU'}$ by $\hat{\mathcal{Y}}_{PU'}$, $(\mathcal{Y}_{\dot{P}\dot{U}}^\perp)_{\dot{P}Y'}$ by $\hat{\mathcal{Y}}_{\dot{P}Y'}$, and $(\mathcal{Y}_{\dot{P}\dot{Y}}^\perp)_{PU'}$ by $\hat{\mathcal{Y}}_{PU'}$. Note that $(\hat{\mathcal{Y}}_{WP}^1)_{\dot{W}\dot{P}} \supseteq \hat{\mathcal{Y}}_{\dot{W}\dot{P}}^2$. It is easily verified using Theorem 2 that the conditions of the hypothesis of Theorem 46 are satisfied with $\mathcal{Y}$ symbols replaced by $\hat{\mathcal{Y}}$ symbols. Now we rewrite part (a) in terms of the $\hat{\mathcal{Y}}$ as follows.

$$ (\hat{\mathcal{Y}}_{WP}^1 \oplus \hat{\mathcal{Y}}_{\dot{W}\dot{P}}^2) \leftrightarrow ((\hat{\mathcal{Y}}_{P\dot{P}U'Y'} \cap \hat{\mathcal{Y}}_{PU'}) \circ P \dot{P}) = (\hat{\mathcal{Y}}_{W\dot{W}U'Y'} \cap \hat{\mathcal{Y}}_{U'}) \circ W \dot{W}. $$

We therefore have $[(\hat{\mathcal{Y}}_{WP}^1 \oplus \hat{\mathcal{Y}}_{\dot{W}\dot{P}}^2) \leftrightarrow ((\hat{\mathcal{Y}}_{P\dot{P}U'Y'} \cap \hat{\mathcal{Y}}_{PU'}) \circ P \dot{P})]^{\perp} = [(\hat{\mathcal{Y}}_{W\dot{W}U'Y'} \cap \hat{\mathcal{Y}}_{WU'}) \circ W \dot{W}]^{\perp}$.

By Theorem 2, we have $(\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \perp \leftrightarrow ((\mathcal{Y}_{P\dot{P}U'Y'} \cap \mathcal{Y}_{PU'}) \circ P \dot{P})^{\perp} = [(\mathcal{Y}_{W\dot{W}U'Y'} \cap \mathcal{Y}_{WU'}) \circ W \dot{W}]^{\perp}$.

$$ \text{Therefore } [(\hat{\mathcal{Y}}_{WP}^1)^{\perp} \oplus (\hat{\mathcal{Y}}_{\dot{W}\dot{P}}^2)^{\perp}] \leftrightarrow [((\hat{\mathcal{Y}}_{P\dot{P}U'Y'})^{\perp} + (\hat{\mathcal{Y}}_{PU'})^{\perp}) \times P \dot{P}] = [\hat{\mathcal{Y}}_{W\dot{W}U'Y'}^{\perp} + \hat{\mathcal{Y}}_{WU'}^{\perp}] \times W \dot{W}. $$

By replacing $P$ with $-\dot{P}$, $\dot{P}$ with $P$, $W$ with $-\dot{W}$, $\dot{W}$ with $W$, $U'$ with $-Y$, and $Y'$ with $U$, we now have the required result. $\square$

**Problem 60.** Prove the following result which shows how controlled and conditioned invariant spaces transfer from the emulator to the original dynamical system through the $\mathcal{E}$-linkage.

**Theorem 49.** Let $\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}}$ be mutual emulators, $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$.

1. Let $\mathcal{Y}_W$ be controlled invariant in $\mathcal{Y}_{W\dot{W}}$. Then $\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_W, (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{WP} \leftrightarrow \mathcal{Y}_W$ are controlled invariant in $\mathcal{Y}_{P\dot{P}}$.
2. Let $\mathcal{V}_W$ be conditioned invariant in $\mathcal{V}_{W\dot{W}}$. Then $\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_W$, $(\mathcal{V}_{\dot{W}\dot{P}}^2)_{WP} \leftrightarrow \mathcal{V}_W$ are conditioned invariant in $\mathcal{V}_{P\dot{P}}$.

Answer

The proof of the theorem needs Theorem 3 and the following definition and lemma extracted from Subsection 5.4 for the convenience of the reader.

**Definition 50.** Let $\mathcal{V}_{AA'}$ be a vector space on $A \uplus A'$ where $A'$ is a disjoint copy of $A$. We say $\mathcal{V}_{AA'}$ is symmetric iff it satisfies the conditions

* if $f_A \in \mathcal{V}_{AA'} \circ A$, then $(f_A, f_{A'}) \in \mathcal{V}_{AA'}$,

* if $(f_A, g_{A'}) \in \mathcal{V}_{AA'}$ then $(g_A, f_{A'}) \in \mathcal{V}_{AA'}$.

**Lemma 51.** Let $\mathcal{V}_{AA'}$ be a symmetric vector space on $A \uplus A'$ where $A'$ is a disjoint copy of $A$. We then have the following.

1. Let $A'$ be a copy of $A$ disjoint from $A \uplus B$. Then $\mathcal{V}_{AA'} \leftrightarrow \mathcal{V}_{AB} = (\mathcal{V}_{AB})_{A'B}$ iff $\mathcal{V}_{AA'} \circ A \supseteq \mathcal{V}_{AB} \circ A$ and $\mathcal{V}_{AA'} \times A \subseteq \mathcal{V}_{AB} \times A$.

2. Let $A'$ be a copy of $A$ disjoint from $A \uplus P \uplus B$. Then $(\mathcal{V}_{AP} \leftrightarrow (\mathcal{V}_{AP})_{A'P}) \leftrightarrow \mathcal{V}_{AB} = (\mathcal{V}_{AB})_{A'B}$ if

$$ \mathcal{V}_{AP} \circ A \supseteq \mathcal{V}_{AB} \circ A, \quad \mathcal{V}_{AP} \times A \subseteq \mathcal{V}_{AB} \times A. $$

For proof see Subsection 5.4

*Proof.* (of Theorem 49) 1. We are given that $\mathcal{V}_{W\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}} \supseteq \mathcal{V}_W$. We will first show that

$$ ((\mathcal{V}_{WP}^1 \oplus \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow (\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_W)_{\dot{P}} \supseteq (\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_W). \quad (53) $$

We will rewrite the LHS of Equation 53 as

$((\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow (\mathcal{V}_{\dot{W}\dot{P}}^1 \leftrightarrow \mathcal{V}_{\dot{W}})$. where $\mathcal{V}_{\dot{W}\dot{P}}^1 \equiv (\mathcal{V}_{WP}^1)_{\dot{W}\dot{P}}$ and $\mathcal{V}_{\dot{W}} \equiv (\mathcal{V}_W)_{\dot{W}}$. $\dot{W}$ being a copy of $\dot{W}$ disjoint from $W \uplus \dot{W} \uplus P \uplus \dot{P}$.

Observe that in this expression no index set appears more than twice. Therefore, by Theorem 3, the expression can have only a unique meaning even without brackets and we can rearrange the terms according to convenience. We now rearrange terms and write the expression as

$$ ((\mathcal{V}_{\dot{W}\dot{P}}^1 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow \mathcal{V}_{\dot{W}} \leftrightarrow \mathcal{V}_{WP}^1. $$

We observe that $(\mathcal{V}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}} = (\mathcal{V}_{W\dot{W}})_{W\dot{W}}$, by Lemma 51 and further that $\mathcal{V}_{\dot{W}\dot{P}}^1 \supseteq \mathcal{V}_{\dot{W}\dot{P}}^2$.

Therefore $(\mathcal{V}_{\dot{W}\dot{P}}^1 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}} \supseteq (\mathcal{V}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}} = (\mathcal{V}_{W\dot{W}})_{W\dot{W}}$. By the controlled invariance of $\mathcal{V}_W$ in $\mathcal{V}_{W\dot{W}}$, we have that $(\mathcal{V}_{W\dot{W}})_{W\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}} \supseteq \mathcal{V}_W$.

Hence $((\mathcal{V}_{\dot{W}\dot{P}}^1 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow \mathcal{V}_{\dot{W}} \leftrightarrow \mathcal{V}_{WP}^1) \supseteq ((\mathcal{V}_{W\dot{W}})_{W\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}}) \leftrightarrow \mathcal{V}_{WP}^1 \supseteq \mathcal{V}_W \leftrightarrow \mathcal{V}_{WP}^1$. This

proves the required result.

We will next show that

$$ ((\mathcal{V}_{WP}^1 \oplus \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow (\mathcal{V}_{WP}^2 \leftrightarrow \mathcal{V}_W)_{\dot{P}} \supseteq (\mathcal{V}_{WP}^2 \leftrightarrow \mathcal{V}_W), \quad (54) $$

where $\mathcal{V}_{WP}^2 \equiv (\mathcal{V}_{WP}^2)_{WP}$.

The LHS of Equation 54 can be rewritten as $((\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow (\mathcal{V}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{V}_{\dot{W}})$. Rearranging terms (using Theorem 3, since no index set appears more than twice), this reduces to $((\mathcal{V}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{V}_{W\dot{W}}) \leftrightarrow \mathcal{V}_{WP}^1 \supseteq \mathcal{V}_W$.
This simplifies, using Lemma 51, to $(\mathcal{Y}_{W\dot{W}}^{1})_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}}^{1} \leftrightarrow \mathcal{Y}_{WP}^{1}$.

We note that $\mathcal{Y}_{WP}^{1} \supseteq (\mathcal{Y}_{\dot{W}\dot{P}}^{2})_{WP}$. Hence

$$((\mathcal{Y}_{\dot{W}\dot{P}}^{1}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^{2}) \leftrightarrow \mathcal{Y}_{W\dot{W}}^{1}) \leftrightarrow \mathcal{Y}_{WP}^{1}) \supseteq ((\mathcal{Y}_{W\dot{W}}^{1})_{\dot{W}\dot{P}} \leftrightarrow \mathcal{Y}_{\dot{W}}^{1}) \leftrightarrow \mathcal{Y}_{WP}^{2} \supseteq \mathcal{Y}_{W} \leftrightarrow \mathcal{Y}_{WP}^{2}.$$

This proves the required result.

(Theorem 49)2. This is dual to the previous result.

We need the following theorems for the proof of the dual (see [19]).

**Theorem 52.** 1. Let $\mathcal{Y}_{P\dot{P}UY} = (\mathcal{Y}_{WP}^{1} \oplus \mathcal{Y}_{\dot{W}\dot{P}}^{2}) \leftrightarrow \mathcal{Y}_{W\dot{W}UY}$. Then $((\mathcal{Y}_{WP}^{1})^{\perp})_{\dot{W}-\dot{P}} \subseteq (\mathcal{Y}_{\dot{W}-\dot{P}}^{2})^{\perp}$ and

$$\mathcal{Y}_{P\dot{P}U'Y'}^{a} = ((\mathcal{Y}_{WP}^{1})_{\dot{W}-\dot{P}}^{\perp} \oplus (\mathcal{Y}_{\dot{W}-\dot{P}}^{2})_{WP}^{\perp}) \leftrightarrow \mathcal{Y}_{W\dot{W}U'Y'}^{a}.$$

$$2. \text{ Let } \mathcal{Y}_{P\dot{P}} = (\mathcal{Y}_{WP}^{1} \oplus \mathcal{Y}_{\dot{W}\dot{P}}^{2}) \leftrightarrow \mathcal{Y}_{W\dot{W}}. \text{ Then } \mathcal{Y}_{P\dot{P}}^{a} = ((\mathcal{Y}_{WP}^{1})_{\dot{W}-\dot{P}}^{\perp} \oplus (\mathcal{Y}_{\dot{W}-\dot{P}}^{2})_{WP}^{\perp}) \leftrightarrow \mathcal{Y}_{W\dot{W}}^{a}.$$

**Theorem 53.** $\mathcal{Y}_W$ is conditioned (controlled) invariant in $\mathcal{Y}_{W\dot{W}}^a$ iff $\mathcal{Y}_{\dot{W}}^\perp$ is controlled (conditioned) invariant in $\mathcal{Y}_{W\dot{W}}^a$.

(proof of the dual continues) By Theorem 52, we know that $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_{P\dot{P}}$ are mutual emulators, $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1$, $\mathcal{Y}_{\dot{W}\dot{P}}^2$. iff $\mathcal{Y}_{W\dot{W}}^a$, $\mathcal{Y}_{P\dot{P}}^a$ are mutual emulators, $\mathcal{E}$-linked through $((\mathcal{Y}_{\dot{W}\dot{P}}^2)^{\perp})_{W-P}$, $((\mathcal{Y}_{WP}^1)^{\perp})_{\dot{W}-\dot{P}}$.

By (Theorem 49)1 above if $\mathcal{Y}_{\dot{W}}^\perp$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}^a$ then $((\mathcal{Y}_{WP}^1)^{\perp})_{W-P} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp, (\mathcal{Y}_{\dot{W}\dot{P}}^2)^{\perp})_{\dot{W}-\dot{P}} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp$ are controlled invariant in $\mathcal{Y}_{P\dot{P}}^a$.

By Theorem 53, $\mathcal{Y}_{\dot{W}}^\perp$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}^a$ iff $\mathcal{Y}_W$ is conditioned invariant in $(\mathcal{Y}_{W\dot{W}}^a)^a = \mathcal{Y}_{W\dot{W}}$, $((\mathcal{Y}_{WP}^1)^{\perp})_{W-P} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp, (\mathcal{Y}_{\dot{W}\dot{P}}^2)^{\perp})_{\dot{W}-\dot{P}} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp$ are controlled invariant in $\mathcal{Y}_{P\dot{P}}^a$, iff $((\mathcal{Y}_{WP}^1)^{\perp})_{W-P} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp)^{\perp} = \mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{W}, ((\mathcal{Y}_{\dot{W}\dot{P}}^2)^{\perp})_{\dot{W}-\dot{P}} \leftrightarrow \mathcal{Y}_{\dot{W}}^\perp)^{\perp} = (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{WP} \leftrightarrow \mathcal{Y}_W$ are conditioned invariant in $\mathcal{Y}_{P\dot{P}}^a$.

Therefore part 2 follows from part 1. $\square$

## 14. Generalized autonomous systems (genauts) and operators (genops)

\* (This section is largely self contained. Motivation is provided by Section 12.)

A generalized autonomous system (genaut) $\mathcal{Y}_{W\dot{W}}$ is on the set $W \oplus \dot{W}$, where $W, \dot{W}$ are disjoint copies of each other. It is called a generalized operator (genop) iff, additionally, it satisfies the conditions

$$(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}, \quad (\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}.$$

### 14.1. Motivation

**Problem 61.** What is the motivation for the notions of genauts and genops, what constraints on operators are relaxed in the case of these notions, and why?

**Answer**

An ‘ordinary’ operator $A$ which takes $w^T$ to $\dot{w}^T$ through $\dot{w}^T = w^T A$ can be treated as the space of all $(w^T, \dot{w}^T)$ which satisfy the equation and can be seen to be a genop. Genauts and genops differ from ordinary operators in the following respects.

* The domain for an operator can be regarded as $\mathcal{F}_W$, the collection of all vectors on $W$. For the more general case of genauts and genops, the role of the domain is played by $\mathcal{Y}_{W\dot{W}} \circ W$.
• In the case of an operator, the range is a subspace of the domain. The analogous idea in the case of the genop is the condition $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$.

• In the case of an operator, $0_W$ maps to $0_{\dot{W}}$. The analogous idea in the case of the genop is $(\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. This has essentially the same 'power' as $0_W$ mapping to $0_{\dot{W}}$.

Let the dynamical system $\mathcal{Y}_{W\dot{W}UY}$ is defined by an equation of the kind

$$ \begin{bmatrix} S_{\dot{W}} & S_W & S_U & S_Y \end{bmatrix} \begin{pmatrix} \dot{w} \\ w \\ u \\ y \end{pmatrix} = 0. \qquad (55) $$

If this system satisfies certain conditions (see Definition 77 of regular dynamical systems), the genaut defined by $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times W\dot{W}$ would be a genop. The linear electrical network in Figure 8 with resistors, inductors and capacitors is of this kind, the variables $w$ being capacitor voltages and inductor currents. Since capacitor loops and inductor cutsets are present, there are constraints on capacitor voltages and inductor currents so that $\mathcal{Y}_{W\dot{W}} \circ W \neq \mathcal{F}_W$. Topological constraints of the kind $Kw = 0$ will result in the constraints $K\dot{w} = 0$, by differentiation. Further, there may be network constraints on capacitor currents and inductor voltages which result in additional constraints on $\dot{w}$. Therefore, in general, $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$. One would like the adjoints of genops to be genops. Therefore we must have $(\mathcal{Y}_{W\dot{W}}^a \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}}^a \circ \dot{W}$, i.e., $((\mathcal{Y}_{W\dot{W}}^{\perp})_{(-\dot{W})W} \circ W)_{\dot{W}} \supseteq (\mathcal{Y}_{W\dot{W}}^{\perp})_{(-\dot{W})W} \circ \dot{W}$, i.e., $((\mathcal{Y}_{W\dot{W}})_{(-\dot{W})W} \times W)_{\dot{W}} \subseteq (\mathcal{Y}_{W\dot{W}})_{(-\dot{W})W} \times \dot{W}$, i.e., $\mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}}$.

**Definition 54.** Let $\mathcal{Y}_{W\dot{W}}$ be a genaut. We say $\mathcal{Y}_W$ is **invariant** in $\mathcal{Y}_{W\dot{W}}$ iff it is both controlled and conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, i.e., iff $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$ and $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$.

## 14.2. Upper and lower semi genops (USG and LSG)

**Problem 62.** An upper semi genop (USG) $\mathcal{Y}_{W\dot{W}}$ satisfies $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$, while a lower semi genop (LSG) $\mathcal{Y}_{W\dot{W}}$ satisfies $(\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. Motivate these definitions by showing that if we start with a dynamical system defined by state-input-output equations and keep input and output free, then we would get a USG, while if we set both input and output to zero, then we would get an LSG.

### Answer

Let $\mathcal{Y}_{W\dot{W}UY}$ be defined by $\dot{w} = Aw + Bu, y = Cw$. If we keep input and output free, i.e., $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \leftrightarrow \mathcal{F}_U \oplus \mathcal{F}_Y$, then since $\mathcal{Y}_{W\dot{W}} \circ W = \mathcal{F}_W$, it is clear that we have $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$. But $\mathcal{Y}_{W\dot{W}} \times \dot{W} = \mathcal{B} := \mathrm{Im}(B)$, while $\mathcal{Y}_{W\dot{W}} \times W = \mathrm{Ker}(A) \cap \mathrm{Ker}(C)$, so that the condition $$(\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}'$$ may not be satisfied.

On the other hand if we set input and output to zero, we have $\mathcal{Y}_{W\dot{W}} \times \dot{W} = \mathbf{0}_{\dot{W}}$, so that $(\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. But $\mathcal{Y}_{W\dot{W}} \circ W = \mathrm{Ker}(C)$ and $\mathcal{Y}_{W\dot{W}} \circ \dot{W} = \mathrm{Im}(A)$, so that so that the condition $$(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}'$$ may not be satisfied.

**Problem 63.** Prove the following lemma which states that genauts which are USG(LSG) remain thus when intersected or summed with certain spaces and therefore that genops remain as such when restricted to, or quotiented over, invariant spaces.

**Lemma 55.** Let $\mathcal{Y}_{W\dot{W}}$ be a genaut. We have the following.
1. If $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_W$ is also conditioned invariant in $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$;
   if $\mathcal{Y}_{W\dot{W}}$ is LSG(USG) then $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$ is LSG(USG).

2. If $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_W$ is also controlled invariant in $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$;
   if $\mathcal{Y}_{W\dot{W}}$ is LSG(USG) then $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$ is LSG(USG).

Answer

Proof. 1. Let $\mathcal{Y}_W$ be conditioned invariant in $\mathcal{Y}_{W\dot{W}}$. Then, $\mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) = \mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}} \subseteq (\mathcal{Y}_W)_{\dot{W}}$,
so that $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$.

Now let $\mathcal{Y}_{W\dot{W}}$ be LSG i.e., $(\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \subseteq \mathcal{Y}_{W\dot{W}} \times W$.

We have $(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \times W = (\mathcal{Y}_{W\dot{W}} \times W) \cap \mathcal{Y}_W$ (see part 2 of Problem 2)

$$ \supseteq (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \cap \mathcal{Y}_W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \text{ (since } \mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq \mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}} \subseteq (\mathcal{Y}_W)_{\dot{W}} \text{)} $$

$$ \supseteq ((\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \times \dot{W})_W. \text{ Thus } \mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W \text{ is LSG.} $$

Next, let $\mathcal{Y}_{W\dot{W}}$ be USG.

$$ \begin{aligned} & \text{We have } (\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \circ W = (\mathcal{Y}_{W\dot{W}} \circ W) \cap \mathcal{Y}_W \supseteq (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W \cap \mathcal{Y}_W \supseteq ((\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \circ \dot{W})_W \cap \mathcal{Y}_W \\ & = ((\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \circ \dot{W})_W \text{ (since by conditioned invariance of } \mathcal{Y}_W, \mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) \circ \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}). \end{aligned} $$

2. This follows by duality from the previous part of the present lemma. $\square$

**Problem 64.** For an operator $A$, two important operations in relation to an invariant subspace $\mathcal{Y}_W$ of its domain are that of restriction to the subspace and that of quotienting over the subspace. For a genaut $\mathcal{Y}_{W\dot{W}}$ in relation to an invariant subspace $\mathcal{Y}_W$ in it, show that

* the restriction $A|_{\mathcal{Y}_W}$ corresponds to $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$ and $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}} \cap (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$.

* the quotient operation $A/\mathcal{Y}_W$ corresponds to $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$ and

$$ \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}}). $$

Answer

Since $\mathcal{Y}_W$ is an invariant subspace for $A$, we have $Aw \in \mathcal{Y}_W$ if $w \in \mathcal{Y}_W$, i.e., $A(\mathcal{Y}_W) \subseteq \mathcal{Y}_W$. For the genaut $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_{W\dot{W}} \leftrightarrow \{w\}$ plays the role of $Aw$. The genaut $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$ corresponds to $A|_{\mathcal{Y}_W}$. Since $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}$, and therefore conditioned invariant in it, we have $\mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$, i.e, whenever $w \in \mathcal{Y}_{W\dot{W}} \circ W \cap \mathcal{Y}_W$ and $(w, \dot{w}) \in \mathcal{Y}_{W\dot{W}}$ we also have $\dot{w} \in (\mathcal{Y}_W)_{\dot{W}}$. Thus $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W \subseteq \mathcal{Y}_{W\dot{W}} \cap (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$. Since the reverse containment is clear, we conclude that $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}} \cap (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$.

Since $\mathcal{Y}_W$ is an invariant subspace for $A$, we have $A(w + \mathcal{Y}_W) \subseteq Aw + \mathcal{Y}_W$. For the genaut $\mathcal{Y}_{W\dot{W}}$, this corresponds to the statement $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\{w\} + \mathcal{Y}_W) \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \{w\}) + (\mathcal{Y}_W)_{\dot{W}}$, $w \in \mathcal{Y}_{W\dot{W}} \circ W$. This latter statement is true because $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\{w\} + \mathcal{Y}_W) = (\mathcal{Y}_{W\dot{W}} \leftrightarrow \{w\}) + \mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_{W\dot{W}} \leftrightarrow \{w\}) + (\mathcal{Y}_W)_{\dot{W}}$. (For collections of vectors $\mathcal{K}_W^1, \mathcal{K}_W^2 \subseteq \mathcal{Y}_{W\dot{W}} \circ W$, it can be verified that $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{K}_W^1 + \mathcal{K}_W^2) = \mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{K}_W^1 + \mathcal{Y}_{W\dot{W}} \leftrightarrow \mathcal{K}_W^2$ (see part 2(b) of Problem 2).)

We define the quotient operator $A/\mathcal{Y}_W$, $\mathcal{Y}_W$ invariant under $A$, by $A/\mathcal{Y}_W(w + \mathcal{Y}_W) := Aw + \mathcal{Y}_W$ modulo $\mathcal{Y}_W$. For the genaut $\mathcal{Y}_{W\dot{W}}$, let us define $\mathcal{Y}_{W\dot{W}}^q := \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$. We then have $\mathcal{Y}_{W\dot{W}}^q \leftrightarrow \{w + \mathcal{Y}_W\} = (\mathcal{Y}_{W\dot{W}} \leftrightarrow \{w\}) + (\mathcal{Y}_W)_{\dot{W}}$ and thus $\mathcal{Y}_{W\dot{W}}^q$ corresponds to $A/\mathcal{Y}_W$ if $\mathcal{Y}_{W\dot{W}}$ corresponds to the operator $A$.

To see that $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$, we merely have to show that RHS is contained in LHS, since the reverse containment is clear. Let $(w_1, \dot{w}_1) \in \mathcal{Y}_{W\dot{W}}$, $(w_2, \dot{w}_2) \in (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$, so that $(w_1 + w_2, \dot{w}_1 + \dot{w}_2) \in \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$. Since $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}$, and therefore controlled invariant in it, we must have $\mathcal{Y}_{W\dot{W}} \leftrightarrow (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W$. Therefore, there exists $\dot{w}'_2 \in (\mathcal{Y}_W)_{\dot{W}}$ such that
($w_2, \dot{w}_2'$)) ∈ ℛ_{W\dot{W}} ⊆ ℛ_{W\dot{W}} + ( ℛ_W )_{\dot{W}} . Now \dot{w}_2 - \dot{w}_2' ∈ ( ℛ_W )_{\dot{W}} so that (0_W, \dot{w}_2 - \dot{w}_2') ∈ ℛ_{W\dot{W}} + ( ℛ_W )_{\dot{W}} .
Therefore
(w_1, \dot{w}_1) + (w_2, \dot{w}_2') + (0_W, \dot{w}_2 - \dot{w}_2') = (w_1 + w_2, \dot{w}_1 + \dot{w}_2) ∈ ℛ_{W\dot{W}} + ( ℛ_W )_{\dot{W}} .

**Problem 65.** Let $\mathcal{Y}_{W\dot{W}UY}^u$ be the dynamical system defined by $\dot{w}=Aw+Bu, y=Cw$. Let $\mathcal{Y}_{W\dot{W}}^u :=$
$\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}$. We have seen in Problem 62, that $\mathcal{Y}_{W\dot{W}}^u$ is USG. Clearly, $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times W\dot{W}$
is a genop being defined by $\dot{w}=Aw$. If we perform state feedback $u=Fw$ on $\mathcal{Y}_{W\dot{W}UY}$ we get a genop
$\mathcal{Y}_{W\dot{W}}^{sf}$ defined by $\dot{w}=(A+BF)w$. Classical multivariable theory tells us that there are two important maps
associated with $\mathcal{Y}_{W\dot{W}}^{sf}$.

1. The first is obtained by restricting the map $A + BF$ to the space $\mathcal{Y}_W$ which is invariant in $\mathcal{Y}_{W\dot{W}}^u$.

2. The second is obtained by quotienting the map $A+BF$ over $\mathcal{Y}_W$. This map depends only on $\mathcal{Y}_{W\dot{W}}^u$ and does not change under state feedback.

Show that the following result is a translation of these ideas into the notation of this section.

**Lemma 56.** Let $\mathcal{Y}_{W\dot{W}}^u$ be USG and let $\mathcal{Y}_{W\dot{W}}^{sf} \subseteq \mathcal{Y}_{W\dot{W}}^u$ be a genop such that $\mathcal{Y}_{W\dot{W}}^u \circ W = \mathcal{Y}_{W\dot{W}}^{sf} \circ W$. Let $\mathcal{Y}_W$ be invariant in $\mathcal{Y}_{W\dot{W}}^u$. Then $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}^{sf}$, the spaces $\mathcal{Y}_{W\dot{W}}^{sf} \cap \mathcal{Y}_W, \mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}}$ are genops and $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$.

Answer

The first is the genaut $\mathcal{Y}_{W\dot{W}}^{sf} \cap \mathcal{Y}_W$, where $\mathcal{Y}_W$ is an invariant space of $\mathcal{Y}_{W\dot{W}}^u$. Since $\mathcal{Y}_{W\dot{W}}^{sf} \subseteq \mathcal{Y}_{W\dot{W}}^u$, and $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}^u$, it is clear that $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}^{sf}$. Since $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}^u$, we have $\mathcal{Y}_{W\dot{W}}^u \circ W \supseteq \mathcal{Y}_W$. By Theorem 43, since $\mathcal{Y}_{W\dot{W}}^{sf} \circ W = \mathcal{Y}_{W\dot{W}}^u \circ W \supseteq \mathcal{Y}_W$, $\mathcal{Y}_W$ is a controlled invariant space of $\mathcal{Y}_{W\dot{W}}^{sf}$. Since $\mathcal{Y}_{W\dot{W}}^{sf}$ is both USG and LSG, by Lemma 55, we have that $\mathcal{Y}_{W\dot{W}}^{sf} \cap \mathcal{Y}_W$ is a genop.

The second is the genaut $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}}$ where $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}^u$. We will show that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}}$ is a genop and that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$. For simplicity, we use the fact that when $\mathcal{Y}_W$ is invariant in a genaut $\mathcal{Y}_{W\dot{W}}', \mathcal{Y}_{W\dot{W}}' + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}' + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ (see Problem 64).

Since $(\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \circ W = \mathcal{Y}_{W\dot{W}}^{sf} \circ W + \mathcal{Y}_W \supseteq (\mathcal{Y}_{W\dot{W}}^{sf} \circ \dot{W} + (\mathcal{Y}_W)_{\dot{W}})_W = ((\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \circ \dot{W})_W$, it is clear that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ is USG. Next, for any genaut $\mathcal{Y}_{W\dot{W}}'$, we must have $(\mathcal{Y}_{W\dot{W}}' + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \times W = \mathcal{Y}_{W\dot{W}}' \times W + \mathcal{Y}_W$ and $(\mathcal{Y}_{W\dot{W}}' + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \times \dot{W} = \mathcal{Y}_{W\dot{W}}' \times \dot{W} + (\mathcal{Y}_W)_{\dot{W}}$. Therefore $(\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \times W = \mathcal{Y}_{W\dot{W}}^{sf} \times W + \mathcal{Y}_W$ and $(\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \times \dot{W} = \mathcal{Y}_{W\dot{W}}^{sf} \times \dot{W} + (\mathcal{Y}_W)_{\dot{W}}$. Since $\mathcal{Y}_{W\dot{W}}^{sf}$ is LSG, it is clear that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ is LSG and since we have already seen that it is USG, $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ is a genop.

It is clear that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}}) \subseteq \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$, since $\mathcal{Y}_{W\dot{W}}^{sf} \subseteq \mathcal{Y}_{W\dot{W}}^u$. We will show that $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$. Let $(fw, g'_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}^u$. Since $\mathcal{Y}_{W\dot{W}}^u \circ W = \mathcal{Y}_{W\dot{W}}^{sf} \circ W$, there exists $(fw, g'_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}^{sf} \subseteq \mathcal{Y}_{W\dot{W}}^u$. Therefore $g'_{\dot{W}} - g'_{\dot{W}} \in \mathcal{Y}_{W\dot{W}}^u \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}}^u \leftrightarrow \mathcal{Y}_W \subseteq (\mathcal{Y}_W)_{\dot{W}}$ since $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}^u$. Therefore $(fw, g'_{\dot{W}}) + (0_W, g'_{\dot{W}} - g'_{\dot{W}}) = (fw, g'_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}}$. Thus $\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$ so that $(\mathcal{Y}_{W\dot{W}}^{sf} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})) \supseteq \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$.

The above also constitutes a proof of Lemma 56.
**Problem 66.** We discuss here the dual situation to the one in Problem 65.

Let $\mathcal{Y}_{W\dot{W}UY}^l$ be the dynamical system defined by $\dot{w}=Aw+Bu, y=Cw$. Let $\mathcal{Y}_{W\dot{W}}^l := \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}$.
We have seen in Problem 62 that $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$ is LSG. If one performs output injection $\dot{w}=Gy$ on this system
we will get a system $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$ governed by $\dot{w}=(A-GC)w$. Clearly, $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$ is a genop. Classical multivariable
theory tells us that there are two important maps associated with $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$.

1. The first is obtained by quotienting the map $A-GC$ over the unobservable space $\mathcal{Y}_W$ which is invariant in $\mathcal{Y}_{W\dot{W}}^l$.

2. The second is a map obtained by restricting the map $A-GC$ to $\mathcal{Y}_W$. This map depends only on $\mathcal{Y}_{W\dot{W}}^l$
and does not change under output injection.

Show that the following result, which is the dual of Lemma 56 and can be proved working with the adjoint
as in Problem 47, is a translation of these ideas into the notation of this section.

**Lemma 57.** Let $\mathcal{Y}_{W\dot{W}}^l$ be LSG and let $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \supseteq \mathcal{Y}_{W\dot{W}}^l$ be a genop such that $\mathcal{Y}_{W\dot{W}}^l \times \dot{W} = \mathcal{Y}_{W\dot{W}}^{\text{oinj}} \times \dot{W}$. Let $\mathcal{Y}_W$ be invariant in $\mathcal{Y}_{W\dot{W}}^l$. Then $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$, the spaces $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} + (\mathcal{Y}_W)_{\dot{W}}$, $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \cap \mathcal{Y}_W$ are genops and $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}}^l \cap \mathcal{Y}_W$.

Answer

The first is the genaut $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}^{\text{oinj}} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$, where $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}^l$.
Since $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \supseteq \mathcal{Y}_{W\dot{W}}^l$ and $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}^l$, it is clear that $\mathcal{Y}_W$ is controlled invariant in
$\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$. Since $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \times \dot{W} = \mathcal{Y}_{W\dot{W}}^l \times \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$, by Theorem 43, $\mathcal{Y}_W$ is a conditioned invariant space of
$\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$. Since $\mathcal{Y}_{W\dot{W}}^{\text{oinj}}$ is both USG and LSG, by Lemma 55, we have that $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} + (\mathcal{Y}_W)_{\dot{W}}$ is a genop.

The second is the genaut $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}}^{\text{oinj}} \cap (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ where $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}^l$. From Lemma 57, it follows that $\mathcal{Y}_{W\dot{W}}^{\text{oinj}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}}^l \cap \mathcal{Y}_W$. Thus this map depends only on $\mathcal{Y}_{W\dot{W}}^l$ and does not change under output injection.

**Problem 67.** Let $\mathcal{Y}_{W\dot{W}U}$ be defined by $\dot{w}=Aw+Bu$, and let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}U} \circ W\dot{W}$. Show that $\mathcal{Y}_{W\dot{W}}$ is USG. Prove the following lemma and use it to modify Algorithm I to show that the minimal conditioned invariant space containing $\text{Im}(B)$ is the column space of $(B|AB|\cdots A^{k-1}B)$, where $A$ is an $n \times n$ matrix ([12]).

**Lemma 58.** Let $\mathcal{Y}_{W\dot{W}}$ be USG and let $\mathcal{Y}_{W\dot{W}} \circ \dot{W} \supseteq \mathcal{Y}_{\dot{W}}^1$, where $\mathcal{Y}_{\dot{W}}^1$ is a copy of $\mathcal{Y}_W^1$. Let, as in Algorithm I, $\mathcal{Y}_{\dot{W}}^{j+1} := (\mathcal{Y}_{\dot{W}\dot{W}} \leftrightarrow \mathcal{Y}_W^j) + \mathcal{Y}_{\dot{W}}^j$. Then if $\tilde{\mathcal{Y}}_W^1 = \mathcal{Y}_{\dot{W}}^1$, $\tilde{\mathcal{Y}}_{\dot{W}}^{j+1} := (\mathcal{Y}_{\dot{W}\dot{W}} \leftrightarrow \tilde{\mathcal{Y}}_W^j)$, we must have, $\sum_{j=1}^k \tilde{\mathcal{Y}}_W^j = \mathcal{Y}_W^k$.

Answer

We have $\mathcal{Y}_{W\dot{W}} \circ W = \mathcal{F}_W$. Therefore $\mathcal{Y}_{W\dot{W}}$ is USG.

**Algorithm III**

Algorithm for computation of unique minimal space containing $\mathcal{Y}_W^{\text{small}}$ and satisfying the condition
($\mathcal{Y}_{WWM} \leftrightarrow \mathcal{Y}_M) \leftrightarrow \tilde{\mathcal{Y}}_W \subseteq \tilde{\mathcal{Y}}_{\dot{W}}$.

Assumption: Let $\mathcal{Y}_{W\dot{W}}$ be USG and let $\mathcal{Y}_{W\dot{W}} \circ \dot{W} \supseteq \mathcal{Y}_{\dot{W}}^{\text{small}}$, where $\mathcal{Y}_{\dot{W}}^{\text{small}}$ is a copy of $\mathcal{Y}_W^{\text{small}}$.

1. Set $\tilde{\mathcal{Y}}_W^1 := \mathcal{Y}_W^{\text{small}}$
2. $\tilde{\gamma}_{\dot{W}}^{j+1} := (\gamma_{\dot{W}W} \leftrightarrow \tilde{\gamma}_{W}^{j})$, $\tilde{\gamma}_{W}^{j+1} := (\tilde{\gamma}_{\dot{W}}^{j+1})_{W}$

3. Stop when

$$ \sum_{j=1}^{k} \tilde{\gamma}_{\dot{W}}^{j} = \sum_{j=1}^{k+1} \tilde{\gamma}_{\dot{W}}^{j}. $$

4. The desired space is $\sum_{j=1}^{k} \tilde{\gamma}_{W}^{j}$.

Let $\mathcal{Y}_{W\dot{W}U}$ be defined by $\dot{w}=Aw+Bu$, and let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}U} \circ W\dot{W}$. In the above algorithm $\tilde{\gamma}_{W}^{1} = \gamma_{W}^{small} = \mathrm{Im}(B)$, $\tilde{\gamma}_{W}^{j} = \mathrm{Im}(A^{j-1}B)$ and the space $\sum_{j=1}^{k} \tilde{\gamma}_{W}^{j}$ is the column space of $(B|AB|\cdots A^{k-1}B)$, when it equals the column space of $(B|AB|\cdots A^{n-1}B)$, where A is an $n \times n$ matrix.

Lemma 58 justifies Algorithm III by showing that it reduces to Algorithm I when $\mathcal{Y}_{W\dot{W}}$ is USG.

*Proof.* (of Lemma 58) We will use the following fact.

Let $\mathcal{H}_A^1, \mathcal{H}_A^2 \subseteq \mathcal{H}_{AB} \circ A$ and let $\mathcal{H}_A := \mathcal{H}_A^1 + \mathcal{H}_A^2$. Then

$(\gamma_{AB} \leftrightarrow \mathcal{H}_A^1) + (\gamma_{AB} \leftrightarrow \mathcal{H}_A^2) = (\gamma_{AB} \leftrightarrow \mathcal{H}_A)$ (see Problem 2). In the present case, we have $\gamma_{W\dot{W}} \circ \dot{W} \supset \gamma_{\dot{W}}^1$ and since $\gamma_{W\dot{W}}$ is USG, it follows that $\gamma_{W\dot{W}} \circ W \supset (\gamma_{W\dot{W}} \circ \dot{W})_W \supset \gamma_W^1$. Since $\gamma_{\dot{W}}^j = (\gamma_{\dot{W}W} \leftrightarrow \gamma_W^{j-1}) + \gamma_W^{j-1}$ and $\tilde{\gamma}_W^1 = \gamma_W^1, \tilde{\gamma}_W^{j+1} := (\gamma_{W\dot{W}} \leftrightarrow \tilde{\gamma}_W^j)$, we have $\gamma_{W\dot{W}} \circ W \supset \tilde{\gamma}_W^j, \gamma_W^j$ for all j.

Let $\gamma_W^{k-1} := \sum_{j=1}^{k-1} \tilde{\gamma}_W^j$. We will show the equality for k. Now,

$$ \begin{aligned} \gamma_{\dot{W}}^{k} &= (\gamma_{\dot{W}W} \leftrightarrow \gamma_{W}^{k-1}) + \gamma_{\dot{W}}^{k-1} \\ &= [\gamma_{\dot{W}W} \leftrightarrow (\tilde{\gamma}_{W}^{k-1} + \gamma_{W}^{k-2})] + \gamma_{\dot{W}}^{k-1} \quad (\text{follows from definition of } \gamma_{W}^{k-1}) \\ &= (\gamma_{\dot{W}W} \leftrightarrow \tilde{\gamma}_{W}^{k-1}) + (\gamma_{\dot{W}W} \leftrightarrow \gamma_{W}^{k-2}) + \gamma_{\dot{W}}^{k-1} \quad (\text{since } \forall j, \gamma_{W\dot{W}} \circ W \supset \tilde{\gamma}_{W}^{j}, \gamma_{W}^{j}) \\ &= (\gamma_{\dot{W}W} \leftrightarrow \tilde{\gamma}_{W}^{k-1}) + \gamma_{\dot{W}}^{k-1} \quad (\text{noting that } (\gamma_{\dot{W}W} \leftrightarrow \gamma_{W}^{k-2}) + \gamma_{\dot{W}}^{k-2} = \gamma_{\dot{W}}^{k-1}) \\ &= \tilde{\gamma}_{\dot{W}}^{k} + \gamma_{\dot{W}}^{k-1}. \end{aligned} $$

The lemma follows since it is given to be true for $k=1$. $\square$

<figure><img src="image_8.png" /></figure>

**Problem 68.** Let $\mathcal{Y}_{W\dot{W}UY}$ be defined by $\dot{w}=Aw+Bu, y=Cw$ and let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}$. Show that $\mathcal{Y}_{W\dot{W}}$ is LSG. Prove the following lemma and use it to modify Algorithm II to show that the maximal controlled invariant space contained in $\mathrm{Ker}(C)$ is the solution space of

$$ \begin{array}{lclcl} Cw = & 0, & & & \\ CAw = & 0, & & & \\ \vdots & & & & \\ CA^{n-1}w = & 0, & & & \end{array} $$

where A is an $n \times n$ matrix ([12]).

**Lemma 59.** Let $\gamma_{W\dot{W}}$ be LSG and let $\gamma_{W\dot{W}} \times W \subseteq \gamma_W^1$. Let, as in Algorithm II, $\gamma_W^{j+1} := (\gamma_{W\dot{W}} \leftrightarrow \gamma_{\dot{W}}^j) \cap \gamma_W^j$, Then if $\tilde{\gamma}_W^1 := \gamma_W^1, \tilde{\gamma}_W^{j+1} := (\gamma_{W\dot{W}} \leftrightarrow \tilde{\gamma}_{\dot{W}}^j)$, we have $\bigcap_{j=1}^k \tilde{\gamma}_W^j = \gamma_W^k$.
Answer

We have $\mathcal{Y}_{W\dot{W}} \times \dot{W} = \mathbf{0}_{\dot{W}}$. Therefore $\mathcal{Y}_{W\dot{W}}$ is LSG.

**Algorithm IV**

Algorithm for computation of unique maximal space contained in $\mathcal{Y}_W^{\text{big}}$ and satisfying the condition $(\mathcal{Y}_{\dot{W}WM} \leftrightarrow \mathcal{Y}_M) \leftrightarrow \tilde{\mathcal{Y}}_{\dot{W}} \supset \tilde{\mathcal{Y}}_W$.

Assumption: Let $\mathcal{Y}_{W\dot{W}}$ be LSG and let $\mathcal{Y}_W^{\text{Big}} \supseteq \mathcal{Y}_{W\dot{W}} \times W$.

1. Set $\tilde{\mathcal{Y}}_W^1 := \mathcal{Y}_W^{\text{Big}}, \tilde{\mathcal{Y}}_{\dot{W}}^1 := (\tilde{\mathcal{Y}}_W^1)_{\dot{W}}$

2. $\tilde{\mathcal{Y}}_W^{j+1} := (\mathcal{Y}_{W\dot{W}W} \leftrightarrow \tilde{\mathcal{Y}}_{\dot{W}}^j)$

3. Stop when

$$ \bigcap_{j=1}^{k} \tilde{\mathcal{Y}}_{W}^{j} = \bigcap_{j=1}^{k+1} \tilde{\mathcal{Y}}_{W}^{j}. $$

4. The desired space is $\bigcap_{j=1}^{k} \tilde{\mathcal{Y}}_{W}^{j}$.

Let $\mathcal{Y}_{W\dot{W}UY}$ be defined by $\dot{w} = Aw + Bu, y = Cw$ and let $\mathcal{Y}_{W\dot{W}} := \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}$. In the above algorithm $\tilde{\mathcal{Y}}_W^1 = \mathcal{Y}_W^{\text{Big}} = \text{Ker}(C), \tilde{\mathcal{Y}}_W^j = \text{Ker}(CA^{j-1})$ and the space $\bigcap_{j=1}^k \tilde{\mathcal{Y}}_W^j$ is the solution space of

$$ Cw = 0, $$

$$ CAw = 0, $$

$$ \vdots $$

$$ CA^{k-1}w = 0, $$

when it reaches the solution space of these set of equations with $k$ replaced by $n$, where $A$ is an $n \times n$ matrix.

Lemma 59 reduces the above algorithm to Algorithm II, when $\mathcal{Y}_{W\dot{W}}$ is LSG. This lemma is the dual of Lemma 58 and can be proved working with the adjoint as in Problem 47.

**15. Polynomials of generalized autonomous systems and operators**

*(This section is based on Section12, Subsection13.2, Section 14.)*

We remind the reader of the following definitions. A generalized autonomous system (genaut) $\mathcal{Y}_{W\dot{W}}$ is on the set $W \uplus \dot{W}$. It is called a generalized operator (genop) iff, additionally, it satisfies the conditions

$$ (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}, \quad (\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}. $$

Polynomials of operators play a fundamental role in studying dynamical systems. We address ourselves in this section to the problem of defining these notions for genauts and genops and to deriving analogous results involving these notions to those in classical linear algebra. To begin with, for an operator $A$, we must build analogues of $A^0, A^j, \sigma A^j, \sigma A^j + \lambda A^k$ and of the operator $0$ on the same domain as that of $A$ and using these, build the analogue $p(\mathcal{Y}_{W\dot{W}})$ of a general polynomial $p(A)$. The aim is for the polynomial $p(\mathcal{Y}_{W\dot{W}})$, under appropriate definitions of the sum ‘+’ and the product ‘*’, to have properties similar to that of $p(A)$, namely the property $p(A) = p_1(A) + p_2(A)$, whenever $p(s) = p_1(s) + p_2(s)$ and the property
$p(A) = p_1(A) * p_2(A)$, whenever $p(s) = p_1(s) * p_2(s)$. Computing this polynomial should not require first casting the dynamical system (for instance the electrical network of Figure 8) in the state-input-output form and then setting input to zero keeping output free. For eigenvalue ideas to work, we need the idea of minimal annihilating polynomial to go through. This needs the construction of the analogue of annihilation of operators to begin with. We must ensure, in particular, that the minimal annihilating polynomial of $\mathcal{Y}_{W\dot{W}}$ is the same as that of the matrix $A$ where $\mathcal{Y}_{W\dot{W}}$ has the representation $\dot{x}= Ax, w = Cx$. We will show that polynomials of genops do satisfy the 'remainder theorem' so that they have minimal annihilating polynomials. As one can expect, polynomials of genauts do not, in general, satisfy the remainder theorem.

### 15.1. Motivation and justification for the definitions on polynomials of genops

**Problem 69.** Let $\mathcal{Y}_{W\dot{W}}$ be a genop and let $p(s) = s^2 + 2s + 3$. How would you define $p(\mathcal{Y}_{W\dot{W}})$? Motivate your definition. Thence define, for a polynomial $q(s) := \sum_0^n \alpha_i s^i$, the polynomial $q(\hat{\mathcal{Y}}_{W\dot{W}})$ of the genaut $\hat{\mathcal{Y}}_{W\dot{W}}$ to be $q(\hat{\mathcal{Y}}_{W\dot{W}}) := \sum_0^n \alpha_i \dot{\hat{\mathcal{Y}}}^{(i)}_{W\dot{W}}$.

**Answer**

If $A$ is an operator we could think of the corresponding genop as $\mathcal{Y}_{W\dot{W}} := \{(w^T, w^T A), w^T \in \mathcal{F}_W\}$. If $A', A''$ are two operators, corresponding to $\mathcal{Y}'_{W\dot{W}}$, $\mathcal{Y}''_{W\dot{W}}$, respectively, $A' + A''$ corresponds to $\mathcal{Y}'_{W\dot{W}} + \mathcal{Y}''_{W\dot{W}} := \{(w, \dot{w}' + \dot{w}'), (w, \dot{w}') \in \mathcal{Y}'_{W\dot{W}}, (w, \dot{w}'') \in \mathcal{Y}''_{W\dot{W}}\}$. However, in general, for a genop $\mathcal{Y}'_{W\dot{W}}$, it should be noted that $\mathcal{Y}'_{W\dot{W}} \circ W \neq \mathcal{F}_W$ and $\mathcal{Y}'_{W\dot{W}} \times \dot{W} \neq \mathbf{0}_{\dot{W}}$. Also, two genops $\mathcal{Y}'_{W\dot{W}}$, $\mathcal{Y}''_{W\dot{W}}$, which permit the definition of $\mathcal{Y}'_{W\dot{W}} + \mathcal{Y}''_{W\dot{W}}$ may not have the same restriction to $W$.

For an operator $A$, if $p(s) := s^2+2s+3$, we would define $p(A) := A^2+2A+3A^0$. Here $A^2(x) := A(A(x))$. Let $\mathcal{Y}_{W\dot{W}} := \{(w^T, w^T A), w^T \in \mathcal{F}_W\}$. It is clear that $A^2$ has to correspond to $\mathcal{Y}^{(2)}_{W\dot{W}} := \{(w^T, w^T A^2), w^T \in \mathcal{F}_W\}$, i.e., $\mathcal{Y}^{(2)}_{W\dot{W}} := (\mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1\dot{W}}$. Therefore we need to define the product $\mathcal{Y}^1_{W\dot{W}} * \mathcal{Y}^2_{W\dot{W}}$ of two genops $\mathcal{Y}^1_{W\dot{W}}$, $\mathcal{Y}^2_{W\dot{W}}$ to be $(\mathcal{Y}^1_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}^2_{W\dot{W}})_{W_1\dot{W}}$.

The operator $2A$ corresponds to $\{(w^T, 2w^T A), w^T \in \mathcal{F}_W\}$, i.e., to $2^{\dot{w}}\mathcal{Y}_{W\dot{W}} := \{(w^T, 2w_1^T), (w^T, w_1^T) \in \mathcal{Y}_{W\dot{W}}\}$. The above definitions of $\mathcal{Y}^{(2)}_{W\dot{W}}$, $2^{\dot{w}}\mathcal{Y}_{W\dot{W}}$, permit restriction of the genop to $W$ to be the same as that of the original genop $\mathcal{Y}_{W\dot{W}}$ to $W$ and have contraction of the genop to $\dot{W}$ to be the same as that of the original genop $\mathcal{Y}_{W\dot{W}}$ to $\dot{W}$. (This turns out to be important since distributivity of $*$ with respect to $+_{\dot{w}}$ holds only when certain subsets have a containment relation between them.) Care has to be taken in the definition of $\mathcal{Y}^{(0)}_{W\dot{W}}$ which should correspond to $A^0 = I$. Let $I_{W\dot{W}} := \{(w^T, \dot{w}^T) := (w^T, w^T), w \in \mathcal{F}_W\}$. If $\mathcal{Y}_{W\dot{W}} \circ W \neq \mathcal{F}_W$ and $\mathcal{Y}_{W\dot{W}} \times \dot{W} \neq \mathbf{0}_{W}$, we would have $I_{W\dot{W}} \circ W \neq \mathcal{Y}_{W\dot{W}} \circ W$, $I_{W\dot{W}} \times \dot{W} \neq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. We would like $\mathcal{Y}^{(0)}_{W\dot{W}} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$ and $\mathcal{Y}^{(0)}_{W\dot{W}} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$, as in the case of $\mathcal{Y}^{(2)}_{W\dot{W}}$, $2^{\dot{w}}\mathcal{Y}_{W\dot{W}}$. If we define $\mathcal{Y}^{(0)}_{W\dot{W}} := I_{W\dot{W}} + \mathcal{Y}^{(0)}_{W\dot{W}} \circ W \oplus \mathcal{Y}^{(0)}_{W\dot{W}} \times \dot{W}$, this desired condition would hold. We remind the reader that $\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}} := \{(f_W, \lambda f_{\dot{W}}), (f_W, f_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}\} + \mathcal{Y}_{W\dot{W}} \times \dot{W}$ (see Problem 10).

Thus when $p(A) = A^2 + 2A + 3A^0$, we define $p(\mathcal{Y}_{W\dot{W}}) := \mathcal{Y}^{(2)}_{W\dot{W}} + \mathcal{Y}^{(2)}_{W\dot{W}} + 2^{\dot{w}}(\mathcal{Y}_{W\dot{W}}) + \mathcal{Y}^{(0)}_{W\dot{W}}$.

We now give a formal description of polynomials of genauts and genops. For this we require the definition of sum, scalar multiplication and 'operator' multiplication. These are captured respectively by '$+_{\dot{w}}', '$\lambda^{\dot{w}}', $ and '$*$' operations. The '$*$' operation is built in terms of '$\leftrightarrow$' operation and is defined below.

**Definition 60.** Let $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}'_{W\dot{W}}$ be generalized autonomous systems. We define

$$ \mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}} \equiv (\mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}. $$
Note that in the above definition, in the case of $\mathcal{Y}_{W\dot{W}}^1$, the 'dummy' set $W_1$ is treated as a copy of $\dot{W}$, and in the case of $\mathcal{Y}_{W\dot{W}}^2$, as a copy of $W$. This operation is associative, if we introduce disjoint dummy sets whenever required. Consider the expression

$$ \mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2 * \dots * \mathcal{Y}_{W\dot{W}}^{i-1} * \mathcal{Y}_{W\dot{W}}^i * \dots * \mathcal{Y}_{W\dot{W}}^{n-1} * \mathcal{Y}_{W\dot{W}}^n $$

This could be treated as

$$ (\mathcal{Y}_{W\dot{W}}^1)_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1W_2} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^i)_{W_{i-1}W_{i-2}} \dots \leftrightarrow (\mathcal{Y}_{W\dot{W}}^{n-1})_{W_{n-2}W_{n-1}} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^n)_{W_{n-1}\dot{W}} $$

where all the $W_i$ can be treated as disjoint from each other and from each of $W, \dot{W}$. It then follows from Theorem 3, that the meaning of the expression with '*' is independent of where brackets are introduced, i.e., the '*' operation is associative.

We now define powers $\mathcal{Y}_{W\dot{W}}^k$ of a genaut $\mathcal{Y}_{W\dot{W}}$ and thence polynomials of $\mathcal{Y}_{W\dot{W}}$.

**Definition 61.** Let $\mathcal{Y}_{W\dot{W}}$ be a genaut. We define

$$ \mathcal{Y}_{W\dot{W}}^{(0)} = I_{W\dot{W}} + \dot{w} [\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}]. $$

$$ \mathcal{Y}_{W\dot{W}}^{(1)} = \mathcal{Y}_{W\dot{W}} $$

$$ \mathcal{Y}_{W\dot{W}}^{(k)} = (\mathcal{Y}_{W\dot{W}}^{(k-1)}) * (\mathcal{Y}_{W\dot{W}}), \quad k = 2, \dots $$

(By associativity of * operation $\mathcal{Y}_{W\dot{W}}^{(k)} = (\mathcal{Y}_{W\dot{W}}^{(k-j)}) * (\mathcal{Y}_{W\dot{W}}^{(j)}$, $j = 1, 2, \dots, k-1$.)

For a polynomial $q(s) := \sum_0^n \alpha_i s^i$, polynomial $q(\hat{\gamma}_{W\dot{W}})$ $\equiv \sum_0^n \alpha_i \dot{w}^i (\hat{\gamma}_{W\dot{W}}^{(i)})$, where

$$ \sum_{0}^{n} \alpha_{i} \dot{w}(\hat{\gamma}_{W\dot{W}}^{(i)}) = \alpha_{0} \dot{w}(\hat{\gamma}_{W\dot{W}}^{(0)}) + \dot{w} \alpha_{1} \dot{w}(\hat{\gamma}_{W\dot{W}}^{(1)}) + \dot{w} \dots + \dot{w} \alpha_{i} \dot{w}(\hat{\gamma}_{W\dot{W}}^{(i)}) + \dot{w} \dots + \dot{w} \alpha_{n} \dot{w}(\hat{\gamma}_{W\dot{W}}^{(n)}) \text{ and } \hat{\gamma}_{W\dot{W}}^{(1)} := \hat{\gamma}_{W\dot{W}} $$

In general, for a genaut $\mathcal{Y}_{W\dot{W}}$, it would not be true that $\mathcal{Y}_{W\dot{W}} * \mathcal{Y}_{W\dot{W}}^{(0)}$ is equal to $\mathcal{Y}_{W\dot{W}}$. But this would be true for genops (see Problem 70 below), which is adequate for our purposes.

The characteristic properties of $\mathcal{Y}_{W\dot{W}}^{(0)}$, for the most important situation, are listed in the following lemma.

**Lemma 62.** Let $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. Then

1. $(f_W, g_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}^{(0)}$ iff $f_W \in \mathcal{Y}_{W\dot{W}} \circ W$, $g_{\dot{W}} \in (f_{\dot{W}} + \mathcal{Y}_{W\dot{W}} \times \dot{W})$, i.e.,
   iff $g_{\dot{W}} \in (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}}$, $f_W \in (g_{\dot{W}} + \mathcal{Y}_{W\dot{W}} \times \dot{W})_W$.

2. $\mathcal{Y}_{W\dot{W}}^{(0)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W, \mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W} = (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}}, \mathcal{Y}_{W\dot{W}}^{(0)} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W, \mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$.
   Hence $\mathcal{Y}_{W\dot{W}}^{(0)}$ is a genop.

3. $(\mathcal{Y}_{W\dot{W}}^{(0)})^a = (\mathcal{Y}_{W\dot{W}}^a)^{(0)}$.

**Problem 70.** 1 Prove the following.

**Lemma 63.** Let $\mathcal{Y}_{W\dot{W}}'$ be a genop such that $\mathcal{Y}_{W\dot{W}}' \circ W = \mathcal{Y}_{W\dot{W}}' \circ W, \mathcal{Y}_{W\dot{W}}' \times \dot{W} = \mathcal{Y}_{W\dot{W}}' \times \dot{W}$. Then
$$ 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}' * \mathcal{Y}_{W\dot{W}}' = \mathcal{Y}_{W\dot{W}}' * 0^{\dot{w}}\mathcal{Y}_{W\dot{W}} = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}} \text{ and } \mathcal{Y}_{W\dot{W}}^{(0)} * \mathcal{Y}_{W\dot{W}}' = \mathcal{Y}_{W\dot{W}}' * \mathcal{Y}_{W\dot{W}}^{(0)} = \mathcal{Y}_{W\dot{W}}' $$

2 The operators $A^0, \lambda A^k, 0 \times A$ have the relations

$$ A^0 * \lambda A^k = \lambda A^k, 0 \times A^k = 0 \times A, \lambda A^k * 0 \times A = 0 \times A, \lambda A^k + 0 \times A = \lambda A^k, \\ \text{if } p_1(s), p_2(s) \text{ are polynomials in } s, \text{ then} $$

$$ (p_1(A))^{(0)} = A^{(0)}, (0 \times (p_1(A))) * (p_2(A)) = (0 \times (p_2(A))) * (p_1(A)) = 0 \times A. \\ \text{Verify the following analogues of these relations when } \mathcal{Y}_{W\dot{W}} \text{ is a genop.} $$
$$
(a) \mathcal{Y}_{W\dot{W}}^{(0)} * (\lambda \dot{w}(\mathcal{Y}_{W\dot{W}}^{(k)})) = (\lambda \dot{w}(\mathcal{Y}_{W\dot{W}}^{(k)})) * \mathcal{Y}_{W\dot{W}}^{(0)} = (\lambda \dot{w}(\mathcal{Y}_{W\dot{W}}^{(k)})),
$$

$$
(b) 0^{\dot{w}}(\lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)})) = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}},
$$

$$
(c) (\lambda^w(\mathcal{Y}_{W\dot{W}}^{(k)})*0^w\mathcal{Y}_{W\dot{W}} = 0^w\mathcal{Y}_{W\dot{W}}*(\lambda^w(\mathcal{Y}_{W\dot{W}}^{(k)})) = 0^w\mathcal{Y}_{W\dot{W}},
$$

$$
(d) 0^{\dot{w}}\mathcal{Y}_{W\dot{W}} + \dot{w}(\lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)})) = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)}),
$$

$$
\begin{align*}
& \text{(Using Lemma 66 'If } \mathcal{Y}_{W\dot{W}} \text{ is genop, then } p(\mathcal{Y}_{W\dot{W}}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W, p(\mathcal{Y}_{W\dot{W}}) \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}, \\
& \qquad p(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W} \text{ and } p(\mathcal{Y}_{W\dot{W}}) \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W. \text{ Hence } p(\mathcal{Y}_{W\dot{W}}) \text{ is a genop.')} \\
& \text{if } p_1(s), p_2(s) \text{ are polynomials in } s, \text{ then } (p_1(\mathcal{Y}_{W\dot{W}}))^{(0)} = \mathcal{Y}_{W\dot{W}}^{(0)} \\
& \qquad (0^{\dot{w}}(p_1(\mathcal{Y}_{W\dot{W}}))) * (p_2(\mathcal{Y}_{W\dot{W}})) = (0^{\dot{w}}(p_2(\mathcal{Y}_{W\dot{W}}))) * (p_1(\mathcal{Y}_{W\dot{W}})) = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}. \\
& \qquad (0^{\dot{w}}(p_1(\mathcal{Y}_{W\dot{W}}))) + \dot{w}(p_2(\mathcal{Y}_{W\dot{W}})) = (p_2(\mathcal{Y}_{W\dot{W}})).
\end{align*}
$$

3. We define $\mathcal{Y}_{W\dot{W}}^{(0)} := I_{W\dot{W}} + \dot{w}(\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W})$, instead of taking it to be equal to $I_{W\dot{W}} := \{(f_W, f_{\dot{W}}), f_W \in \mathcal{F}_W\}$.

* We define $0^w \mathcal{Y}_{W\dot{W}} := (\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W})$, instead of taking it to be equal to $\mathcal{F}_W \oplus \mathbf{0}_{\dot{W}}$.

Explain why.

4 Show that $\mathcal{Y}_{W\dot{W}}^{(0)}$, $0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$ are genops, if $\mathcal{Y}_{W\dot{W}}$ is USG or LSG.

5 Prove the following

**Lemma 64.** Let $\mathcal{Y}_{W\dot{W}}$ be a genop. Then $\lambda_j^\dot{w}(\mathcal{Y}_{W\dot{W}}^{(j)}) * \lambda_k^\dot{w}(\mathcal{Y}_{W\dot{W}}^{(k)}) = (\lambda_j \lambda_k)^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j+k)})$.

Answer

1. Let $\mathcal{Y}'_{W\dot{W}}$ be a genop such that $\mathcal{Y}'_{W\dot{W}} \circ W = \mathcal{Y}_{W\dot{W}} \circ W, \mathcal{Y}'_{W\dot{W}} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$. We have
$(0^w \mathcal{Y}'_{W\dot{W}}) * \mathcal{Y}'_{W\dot{W}} = (\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W})_{WW_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}} = \mathcal{Y}_{W\dot{W}} \circ W \oplus [(\mathcal{Y}_{W\dot{W}} \times \dot{W})_{W_1}] \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$. Now $(\mathcal{Y}_{W\dot{W}} \times \dot{W})_{W_1} \subseteq (\mathcal{Y}_{W\dot{W}} \circ W)_{W_1} = (\mathcal{Y}'_{W\dot{W}} \circ W)_{W_1}$, so that $(\mathcal{Y}_{W\dot{W}} \times \dot{W})_{W_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}} = (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$. Therefore, $(0^w \mathcal{Y}'_{W\dot{W}}) * \mathcal{Y}'_{W\dot{W}} = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W} = 0^w \mathcal{Y}'_{W\dot{W}}$. The
proof of the fact that $\mathcal{Y}'_{W\dot{W}} * (0^w \mathcal{Y}'_{W\dot{W}}) = 0^w \mathcal{Y}'_{W\dot{W}}$, is similar.

2(a). Let $\mathcal{Y}'_{W\dot{W}}$ be as in part 1 above. We have
$(\mathcal{Y}_{W\dot{W}} \circ W \oplus (\mathcal{Y}'_{W\dot{W}})_{WW_1} \times W_1)) \circ W_1 = (\mathcal{Y}_{W\dot{W}})_{WW_1} \times W_1 \subseteq (\mathcal{Y}_{W\dot{W}} \circ W)_{W_1}$ (since $\mathcal{Y}_{W\dot{W}}$ is a genop)
$= (\mathcal{Y}'_{W\dot{W}} \circ W)_{W_1}$. We can therefore use the distributivity rule in Theorem 18, part 4 and conclude
$((I_{W\dot{W}})_{WW_1} + w_1 ((\mathcal{Y}_{W\dot{W}} \circ W \oplus (\mathcal{Y}_{W\dot{W}})_{WW_1} \times W_1))) \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$
$= ((I_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}) + \dot{w}((\mathcal{Y}_{W\dot{W}} \circ W \oplus (\mathcal{Y}_{W\dot{W}})_{WW_1} \times W_1)) \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$
$= \mathcal{Y}'_{W\dot{W}} + \dot{w}((\mathcal{Y}_{W\dot{W}} \circ W \oplus (\mathcal{Y}_{W\dot{W}})_{WW_1} \times W_1)) \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$
$= \mathcal{Y}'_{W\dot{W}} + \dot{w}((\mathcal{Y}'_{W\dot{W}} \circ W \oplus (\mathcal{Y}'_{W\dot{W}})_{WW_1} \times W_1)) \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$
$\mathcal{Y}'_{W\dot{W}} + \dot{w}(\mathcal{Y}'_{W\dot{W}} \circ W \oplus \mathcal{Y}'_{W\dot{W}} \times \dot{W}) = \mathcal{Y}'_{W\dot{W}}$.
Thus $\mathcal{Y}_{W\dot{W}}^{(0)} * \mathcal{Y}'_{W\dot{W}} = \mathcal{Y}'_{W\dot{W}}$.

We can similarly prove $\mathcal{Y}'_{W\dot{W}} * \mathcal{Y}_{W\dot{W}}^{(0)} = \mathcal{Y}'_{W\dot{W}}$. But here we have to use the distributivity rule in Theorem 18, part 2 with the fact that $\mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}}$.

Parts b,c,d, follow routinely from the relevant definitions. Part (e) is a consequence of the previous parts and part 1.

3. The necessary properties listed in 2. above, hold even with the present definition. We would like to treat $\mathcal{Y}_{W\dot{W}}^{(0)}$, $0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(k)}$, as polynomials of $\mathcal{Y}_{W\dot{W}}$ and for such polynomials when $\mathcal{Y}_{W\dot{W}}$ is a genop, a convenient
result is that their restriction to $W$ and contraction to $\dot{W}$, are respectively equal to $\mathcal{Y}_{W\dot{W}} \circ W$, $\mathcal{Y}_{W\dot{W}} \times \dot{W}$ (Lemma 66). However, this result fails if we define $\mathcal{Y}_{W\dot{W}}^{(0)}$ as $I_{W\dot{W}}$ and $0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$ as $\mathcal{F}_W \oplus \mathbf{0}_{\dot{W}}$.

4. We have for a genaut $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_{W\dot{W}}^{(0)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $\mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$. When $\mathcal{Y}_{W\dot{W}}$ is USG, or if it is LSG, we also have $\mathcal{Y}_{W\dot{W}}^{(0)} \circ W \supset (\mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W})_W$. In these cases, it follows from the definition of $\mathcal{Y}_{W\dot{W}}^{(0)}$, that $\mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W} = (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}}$, and $\mathcal{Y}_{W\dot{W}}^{(0)} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W$. Therefore, $\mathcal{Y}_{W\dot{W}}^{(0)}$ is a genop. Again, when $\mathcal{Y}_{W\dot{W}}$ is USG, or if it is LSG, since $0^{\dot{w}}\mathcal{Y}_{W\dot{W}} = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}$, it follows that $0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$ is a genop.

5. (Proof of Lemma 64) Let $\lambda_j, \lambda_k$ be nonzero. Then $(x_W, \lambda_j\lambda_k z_{\dot{W}}) \in \lambda_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}) * \lambda_k^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)})$ iff there exist $((x_W, \lambda_j y_{\dot{W}}) \in \lambda_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}), (\lambda_j y_W, \lambda_j \lambda_k z_{\dot{W}}) \in \lambda_k^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)}), \text{i.e., iff there exist} \\
(x_W, y_{\dot{W}}) \in (\mathcal{Y}_{W\dot{W}}^{(j)}), (y_W, z_{\dot{W}}) \in (\mathcal{Y}_{W\dot{W}}^{(k)}), \text{i.e., iff there exist} \\
(x_W, z_{\dot{W}}) \in (\mathcal{Y}_{W\dot{W}}^{(j+k)}), \text{i.e., iff } (x_W, \lambda_j \lambda_k z_{\dot{W}}) \in (\lambda_j \lambda_k)^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j+k)}).$ \\
When $\lambda_j$ or $\lambda_k$ equals zero, it follows from part 2(e) of the present problem, that \\
$\lambda_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}) * \lambda_k^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)}) = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}} = (\lambda_j \lambda_k)^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j+k)})$.

## 15.2. Ad joints of polynomials of genauts

**Problem 71.** Let $\mathcal{Y}_{W\dot{W}}$ be a genaut and let $p(s)$ be a polynomial. Show that $(p(\mathcal{Y}_{W\dot{W}}))^{a} = p(\mathcal{Y}_{W\dot{W}}^{a})$.

### Answer

We will show the following, from which the result follows. (Note that $\mathcal{Y}_{W\dot{W}}^{(0)}$ has to be treated as a polynomial in $\mathcal{Y}_{W\dot{W}}$, since it is needed to take care of the constant term. For instance if $p(s) = s^2 + 2s + 3$, we define $p(\mathcal{Y}_{W\dot{W}}) := \mathcal{Y}_{W\dot{W}}^{(2)} + \dot{w}2^{\dot{w}}(\mathcal{Y}_{W\dot{W}}) + \dot{w}3^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(0)})$. For simplicity $p(\mathcal{Y}_{W\dot{W}}) := 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$ should also be permitted as a polynomial in $\mathcal{Y}_{W\dot{W}}$.)

$$1. (0^{\dot{w}}\mathcal{Y}_{W\dot{W}})^a = (0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^a),$$

$$2. (\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2)^a = (\mathcal{Y}_{W\dot{W}}^2)^a * (\mathcal{Y}_{W\dot{W}}^1)^a. \text{ Hence } (\mathcal{Y}_{W\dot{W}}^{(k)})^a = (\mathcal{Y}_{W\dot{W}}^a)^{(k)}, \quad k = 1, \dots,$$

$$3. (\mathcal{Y}_{W\dot{W}}^1 + \dot{w}\mathcal{Y}_{W\dot{W}}^2)^a = (\mathcal{Y}_{W\dot{W}}^1)^a + \dot{w}(\mathcal{Y}_{W\dot{W}}^2)^a.$$

$$4. (\mathcal{Y}_{W\dot{W}}^{(0)})^a = (\mathcal{Y}_{W\dot{W}}^a)^{(0)}.$$

$$5. (\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}})^a = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a). \text{ Hence } (\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(k)})^a = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)})^a = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a)^{(k)}, \quad k = 0, 1, \dots,$$

$$6. (\lambda_i^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(i)} + \dot{w}\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)})^a = \lambda_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a)^{(i)} + \dot{w}\lambda_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a)^{(j)}.$$

$$\begin{aligned}
\text{Proof. 1. } (0^{\dot{w}}\mathcal{Y}_{W\dot{W}})^a &= (\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W})^a = (\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W})^{\perp}_{(-\dot{W})W} \\
&= ((\mathcal{Y}_{W\dot{W}}^{\perp} \times W \oplus \mathcal{Y}_{W\dot{W}}^{\perp} \circ \dot{W})_{(-\dot{W})W}) = ((\mathcal{Y}_{W\dot{W}}^{\perp})_{(-\dot{W})W}) \circ W \oplus (\mathcal{Y}_{W\dot{W}}^{\perp})_{(-\dot{W})W} \times \dot{W} \\
&= (\mathcal{Y}_{W\dot{W}}^a \circ W \oplus \mathcal{Y}_{W\dot{W}}^a \times \dot{W}) = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^a.
\end{aligned}$$

2. We have $(\mathcal{Y}_{W\dot{W}}^2)^a * (\mathcal{Y}_{W\dot{W}}^1)^a = ((\mathcal{Y}_{W\dot{W}}^2)^\perp)_{(-W_1)W} \leftrightarrow ((\mathcal{Y}_{W\dot{W}}^1)^\perp)_{(-\dot{W})W_1}$, where we have taken $W_1$ as a copy of $W, \dot{W}$ disjoint from both of them. By Theorem 2, $((\mathcal{Y}_{W\dot{W}}^2)^\perp)_{(-W_1)W} \leftrightarrow ((\mathcal{Y}_{W\dot{W}}^1)^\perp)_{(-\dot{W})W_1} = ((\mathcal{Y}_{W\dot{W}}^2)_{W_1W} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1)_{(-\dot{W})W_1})^\perp$ \\
$= ((\mathcal{Y}_{W\dot{W}}^1)_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}})_{(-\dot{W})W}^\perp = (\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2)^a.$
3. $(\mathcal{Y}_{W\dot{W}}^1 + \dot{\omega}\mathcal{Y}_{W\dot{W}}^2)^a = (\mathcal{Y}_{W\dot{W}}^1 + \dot{\omega}\mathcal{Y}_{W\dot{W}}^2)^\perp (-\dot{W})W = ((\mathcal{Y}_{W\dot{W}}^1)^\perp + \omega (\mathcal{Y}_{W\dot{W}}^2)^\perp) (-\dot{W})W \text{ (using Theorem 16)} \\
= (\mathcal{Y}_{W\dot{W}}^1)^\perp (-\dot{W})W + \dot{\omega}(\mathcal{Y}_{W\dot{W}}^2)^\perp (-\dot{W})W = (\mathcal{Y}_{W\dot{W}}^1)^a + \dot{\omega}(\mathcal{Y}_{W\dot{W}}^2)^a$.

4. $(\mathcal{Y}_{W\dot{W}}^{(0)})^a = (I_{W\dot{W}} + \dot{\omega} 0^{\dot{w}}\mathcal{Y}_{W\dot{W}})^a = (I_{W\dot{W}})^a + \dot{\omega}(0^{\dot{w}}\mathcal{Y}_{W\dot{W}})^a = I_{W\dot{W}} + \dot{\omega} 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^a = (\mathcal{Y}_{W\dot{W}}^a)^{(0)}.$

5. The case $\lambda = 0$, has already been handled in part 1. above. When $\lambda \neq 0$, we have $(\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}}^a)^a = ((\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}}^a)^\perp)_{(-\dot{W})W} = (\lambda^w(\mathcal{Y}_{W\dot{W}}^a)^\perp)_{(-\dot{W})W}) = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a)$ (using Theorem 17). Hence $(\lambda^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(k)})^a = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)})^a = \lambda^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^a)^{(k)}$, $k = 0, 1, \dots$.

6. This follows from parts 2,3 and 5, above.

□

### 15.3. Remainder theorem for polynomials of genops

**Problem 72.** Show that the following 'remainder' theorem holds for polynomials of genops.

**Theorem 65.** Let $\mathcal{Y}_{W\dot{W}}$ be a genop.

1. Let $p(s) = p_1(s) + q(s)$. Then

$$p(\mathcal{Y}_{W\dot{W}}) = (p_1(\mathcal{Y}_{W\dot{W}})) + \dot{\omega}(q(\mathcal{Y}_{W\dot{W}})).$$

2. Let $p(s) = p_1(s)q(s)$. Then

$$p(\mathcal{Y}_{W\dot{W}}) = (p_1(\mathcal{Y}_{W\dot{W}})) * (q(\mathcal{Y}_{W\dot{W}})).$$

3. Let $\mathcal{Y}_{W\dot{W}}$ be a genop and let $p(s) = p_1(s)q(s) + a(s)$. Then

$$p(\mathcal{Y}_{W\dot{W}}) = ((p_1(\mathcal{Y}_{W\dot{W}})) * (q(\mathcal{Y}_{W\dot{W}}))) + \dot{\omega}a(\mathcal{Y}_{W\dot{W}}).$$

**Answer**

We need the following lemma for the proof of Theorem 65.

**Lemma 66.** Let $\mathcal{Y}_{W\dot{W}}$ be a genop and let $p(s)$ be a polynomial in $s$. We have the following.

1. If $\mathcal{Y}_{W\dot{W}}$ is a USG, then $p(\mathcal{Y}_{W\dot{W}}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W$ and $p(\mathcal{Y}_{W\dot{W}}) \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$. Hence $p(\mathcal{Y}_{W\dot{W}})$ is a USG.

2. If $\mathcal{Y}_{W\dot{W}}$ is an LSG, then $p(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$ and $p(\mathcal{Y}_{W\dot{W}}) \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$. Hence $p(\mathcal{Y}_{W\dot{W}})$ is an LSG.

3. If $\mathcal{Y}_{W\dot{W}}$ is genop, then $p(\mathcal{Y}_{W\dot{W}}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $p(\mathcal{Y}_{W\dot{W}}) \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$, $p(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$ and $p(\mathcal{Y}_{W\dot{W}}) \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$. Hence $p(\mathcal{Y}_{W\dot{W}})$ is a genop.

*Proof.* It can be directly verified that

* When $j \neq 0$ and $\mathcal{Y}_{W\dot{W}}$ is a USG, $\mathcal{Y}_{W\dot{W}}^{(j)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $\mathcal{Y}_{W\dot{W}}^{(j)} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$. When $j \neq 0$ and $\mathcal{Y}_{W\dot{W}}$ is an LSG, $\mathcal{Y}_{W\dot{W}}^{(j)} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$, $\mathcal{Y}_{W\dot{W}}^{(j)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$.

* When $j \neq 0$ and $\mathcal{Y}_{W\dot{W}}$ is a USG, $\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$, When $j \neq 0$ and $\mathcal{Y}_{W\dot{W}}$ is an LSG, $\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$, $\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$,
• When $\mathcal{Y}_{W\dot{W}}$ is a USG, $\mathcal{Y}_{W\dot{W}}^{(0)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $\lambda_0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(0)} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$,
$\mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$, $\lambda_0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$,

When $\mathcal{Y}_{W\dot{W}}$ is an LSG, $\mathcal{Y}_{W\dot{W}}^{(0)} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$, $\mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$,
$\lambda_0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(0)} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$, $\lambda_0^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$,

• When $\mathcal{Y}_{W\dot{W}}$ is a USG, $+_{\dot{w}}\{\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}\} \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, $+_{\dot{w}}\{\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}\} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$.
$\mathcal{Y}_{W\dot{W}}$ is an LSG, $+_{\dot{w}}\{\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}\} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W$, $+_{\dot{w}}\{\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}\} \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$.

While computing $p(\mathcal{Y}_{W\dot{W}})$, we combine terms of the kind $\lambda_j^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}$ through $+_{\dot{w}}$ intersection-sum. The result follows. $\square$

*Proof.* (Proof of Theorem 65) 1. This follows because $+_{\dot{w}}$ operation is associative and commutative (Lemma 13) and because $\lambda_1^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} +_{\dot{w}}\lambda_2^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)} = (\lambda_1 + \lambda_2)^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(j)}$. (Note that when $(\lambda_1 + \lambda_2) = 0$, both the LHS and the RHS equal $\mathcal{Y}_{W\dot{W}}^{(j)} \circ W \oplus \mathcal{Y}_{W\dot{W}}^{(j)} \times \dot{W}$. In the case of the LHS, this follows by definition of $+_{\dot{w}}$ and in the case of the RHS since $0^{\dot{W}}\mathcal{Y}_{W\dot{W}}^{(j)} := \mathcal{Y}_{W\dot{W}}^{(j)} \circ W \oplus \mathcal{Y}_{W\dot{W}}^{(j)} \times \dot{W}$.)

2. The second part requires more work since distributivity of $*$ over $+_{\dot{w}}$ holds only under conditions listed in Theorem 18, rewritten below for convenience of the reader.'

1. If $\mathcal{Y}_{CB} \times B \subseteq \mathcal{Y}_{BA}^{1} \times B, \dots, \mathcal{Y}_{CB} \times B \subseteq \mathcal{Y}_{BA}^{k-1} \times B$, then

$$ \mathcal{Y}_{CB} \leftrightarrow (\mathcal{Y}_{BA}^{1} + a \cdots + a \mathcal{Y}_{BA}^{k}) = (\mathcal{Y}_{CB} \leftrightarrow \mathcal{Y}_{BA}^{1}) + a \cdots + a (\mathcal{Y}_{CB} \leftrightarrow \mathcal{Y}_{BA}^{k}). $$

2. If $\mathcal{Y}_{BC} \circ B \supseteq \mathcal{Y}_{AB}^{1} \circ B, \dots, \mathcal{Y}_{BC} \circ B \supseteq \mathcal{Y}_{AB}^{k-1} \circ B$, then

$$ (\mathcal{Y}_{AB}^{1} + b \cdots + b \mathcal{Y}_{AB}^{k}) \leftrightarrow \mathcal{Y}_{BC} = (\mathcal{Y}_{AB}^{1} \leftrightarrow \mathcal{Y}_{BC}) + c \cdots + c (\mathcal{Y}_{AB}^{k} \leftrightarrow \mathcal{Y}_{BC}). $$

Let $p_1(s) \equiv \sum_{j=0}^{n} \alpha_j s^j$, $q(s) \equiv \sum_{i=0}^{m} \beta_i s^i$. Then

$$ p_1(\mathcal{Y}_{W\dot{W}}) = \sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}), \quad q(\mathcal{Y}_{W\dot{W}}) = \sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}). $$

We are given that $p(s) = p_1(s)q(s)$. In order to show that $p(\mathcal{Y}_{W\dot{W}}) = (p_1(\mathcal{Y}_{W\dot{W}})) * (q(\mathcal{Y}_{W\dot{W}}))$, we will merely show that distributivity holds in the product

$$ (\sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})) * (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})). $$

We have

$$ (p_1(\mathcal{Y}_{W\dot{W}})) * (q(\mathcal{Y}_{W\dot{W}})) = (p_1(\mathcal{Y}_{W\dot{W}})) * (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = (p_1(\mathcal{Y}_{W\dot{W}}))_{W\dot{W}_n} \leftrightarrow (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_n\dot{W}}. $$

Since $\mathcal{Y}_{W\dot{W}}$ is a genop, we have that

$$ (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \subseteq \mathcal{Y}_{W\dot{W}} \times W. $$

From Lemma 66, we have that

$$ p_1(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W} \quad \text{and} \quad \mathcal{Y}_{W\dot{W}} \times W \subseteq \beta_i^{\dot{w}}\mathcal{Y}_{W\dot{W}}^{(i)} \times W, \forall i. $$
It follows that

$$ (p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \times W_n \subseteq \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})_{W_n\dot{W}} \times W_n, \forall i. $$

Therefore by Theorem 18, it follows that

$$ (p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \leftrightarrow (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_n\dot{W}} = \sum_{i=0}^{m} (p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_n\dot{W}} $$

[In expanded form the equation reads

$$ \begin{aligned} (p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} &\leftrightarrow [(\beta_0^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(0)}))_{W_n\dot{W}} + \dot{w} \cdots + \dot{w} (\beta_m^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(m)}))_{W_n\dot{W}}] \\ &= [(p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \leftrightarrow (\beta_0^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(0)}))_{W_n\dot{W}}] + \dot{w} \cdots + \dot{w} [(p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \leftrightarrow (\beta_m^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(m)}))_{W_n\dot{W}}]. \end{aligned} $$

$$ \text{Thus, } (p_1(\mathcal{Y}_{W\dot{W}})) * (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = \sum_{i=0}^{m} (p_1(\mathcal{Y}_{W\dot{W}})) * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})). $$

Next, consider the expression

$$ (p_1(\mathcal{Y}_{W\dot{W}})) * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = (p_1(\mathcal{Y}_{W\dot{W}}))_{WW_n} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_n\dot{W}} = (\sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}))_{WW_n} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_n\dot{W}}. $$

From Lemma 66, we have that

$$ \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}) \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}, \forall j \text{ and } \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W. $$

Since $\mathcal{Y}_{W\dot{W}}$ is a genop, we have that

$$ (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W \subseteq \mathcal{Y}_{W\dot{W}} \circ W. $$

It follows that

$$ \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})_{WW_k} \circ W_k \subseteq (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}} \circ W_k, \forall j. $$

From Theorem 18, we then have that

$$ [\sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})]_{WW_k} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}} = \sum_{j=0}^{n} [(\alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}))_{WW_k}] \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}}. $$

[In expanded form the equation reads

$$ \begin{aligned} & [\alpha_0^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(0)}) + \dot{w} \cdots + \dot{w} \alpha_n^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(n)}))_{WW_k} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}} \\ &= [(\alpha_0^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(0)}))_{WW_k} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}}] + \dot{w} \cdots + \dot{w} [(\alpha_n^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(n)}))_{WW_k} \leftrightarrow (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))_{W_k\dot{W}}]. \end{aligned} $$

Therefore,

$$ (p_1(\mathcal{Y}_{W\dot{W}})) * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = [\sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})] * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = \sum_{j=0}^{n} [(\alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})) * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))]. $$

Thus

$$ (\sum_{j=0}^{n} \alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})) * (\sum_{i=0}^{m} \beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)})) = \sum_{i=0}^{m} \sum_{j=0}^{n} [(\alpha_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)})) * (\beta_i^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(i)}))]. $$

Since by Lemma 64, $\lambda_j^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j)}) * \lambda_k^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(k)}) = (\lambda_j \lambda_k)^{\dot{w}}(\mathcal{Y}_{W\dot{W}}^{(j+k)}),$ it follows that if $p_1(s) := \sum_{j=0}^{n} \alpha_j s^j$, $p_2(s) := \sum_{i=0}^{m} \beta_i s^i$, we have that if $p_1(s) = p_1(s)p_2(s)$, then $p(\mathcal{Y}_{W\dot{W}}) = p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}})$.

3. This is immediate from the two parts above. $\square$
### 15.4. Minimal annihilating polynomial of genops

**Problem 73.**

**Definition 67.** We say a genaut $\mathcal{Y}_{W\dot{W}}$ is decoupled iff it has the form $\mathcal{Y}_W^1 \oplus (\mathcal{Y}_W^2)_{\dot{W}}$. A polynomial $p(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$ iff $p(\mathcal{Y}_{W\dot{W}})$ is decoupled.

1. Show that if the genaut $\mathcal{Y}_{W\dot{W}}$ is decoupled so are $\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}$, $\mathcal{Y}'_{W\dot{W}} * \mathcal{Y}_{W\dot{W}}$.

2. Justify the above definition for the annihilating polynomial of genauts and genops.

**Answer**

1. We will only prove $\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}$ is decoupled.

We have $\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}} := (\mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$ and $\mathcal{Y}_{W\dot{W}} = \mathcal{Y}_W^1 \oplus \mathcal{Y}_{\dot{W}}^2$. Suppose $(x_W, a_{\dot{W}})$, $(y_W, b_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}$. We must then have $(x_W, \alpha_{W_1}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1}$, $(\alpha_{W_1}, a_{\dot{W}}) \in (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$, $(y_W, \beta_{W_1}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1}$, $(\beta_{W_1}, b_{\dot{W}}) \in (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$, for some $\alpha_{W_1}, \beta_{W_1}$. But $\mathcal{Y}_{W\dot{W}} = \mathcal{Y}_W^1 \oplus \mathcal{Y}_{\dot{W}}^2$. Since $(x_W, \alpha_{W_1}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1}$, $(y_W, \beta_{W_1}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1}$, it follows that $(x_W, \beta_{W_1}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1}$, $(\beta_{W_1}, b_{\dot{W}}) \in (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}}$, so that $(x_W, b_{\dot{W}}) \in (\mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}'_{W\dot{W}})_{W_1\dot{W}} = \mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}$.

2. For an operator $A$, we say the polynomial $p(s)$ is an annihilating polynomial iff $p(A) = 0 \times A$. Therefore for genops the natural definition is that $p(s)$ is an annihilating polynomial iff $p(\mathcal{Y}_{W\dot{W}}) = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}} = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}$. The definition works because when $\mathcal{Y}_{W\dot{W}}$ is a genop, $p(\mathcal{Y}_{W\dot{W}}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W, p(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$ (Lemma 66). (Note that when $A$ corresponds to $\mathcal{Y}_{W\dot{W}}$, we have $\mathcal{Y}_{W\dot{W}} \circ W = \mathcal{F}_W, \mathcal{Y}_{W\dot{W}} \times \dot{W} = \mathbf{0}_{\dot{W}}$, if $p(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$, we have $p(\mathcal{Y}_{W\dot{W}}) = \mathcal{F}_W \oplus \mathbf{0}_{\dot{W}}$ which agrees with the operator $0 \times A$.)

However, for a genaut this definition will not work in general. For instance $(\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}) \circ W = \mathcal{Y}_{W\dot{W}} \circ W$, only if $(\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$ and $(\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \times \dot{W}$, only if $(\mathcal{Y}_{W\dot{W}} \times W)_{\dot{W}} \supseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}$. The definition $p(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$ iff $p(\mathcal{Y}_{W\dot{W}})$ is decoupled' captures some of the properties of 'annihilation'. For instance, if either $\mathcal{Y}_{W\dot{W}}^1, \mathcal{Y}_{W\dot{W}}^2$ is decoupled so is $\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2$. So product of an annihilating polynomial and any polynomial of $\mathcal{Y}_{W\dot{W}}$ will appear decoupled. (However, it must be noted that if $p(s) = p_1(s) * p_2(s)$, in general $p(\mathcal{Y}_{W\dot{W}}) \neq p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}})$, for a genaut $\mathcal{Y}_{W\dot{W}}$. Also, for genauts, there is no unique minimal annihilating polynomial.)

When a genaut $\mathcal{Y}_{W\dot{W}}$ is a genop, its annihilating polynomial $p(\mathcal{Y}_{W\dot{W}})$ will be decoupled and be equal to $\mathcal{Y}_W^1 \oplus \mathcal{Y}_{\dot{W}}^2$, for some $\mathcal{Y}_W^1, \mathcal{Y}_{\dot{W}}^2$. But by Lemma 66, this must equal $p(\mathcal{Y}_{W\dot{W}}) \circ W \oplus p(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W} = 0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$. So, for a genop $\mathcal{Y}_{W\dot{W}}$, insisting that the polynomial $p(\mathcal{Y}_{W\dot{W}})$ be decoupled is equivalent to setting it equal to $0^{\dot{w}}\mathcal{Y}_{W\dot{W}}$.

**Problem 74.**

**Definition 68.** We say that a polynomial is a minimal annihilating polynomial for $\mathcal{Y}_{W\dot{W}}$ iff it annihilates $\mathcal{Y}_{W\dot{W}}$ and has the lowest degree $\ge 1$, among all nonzero annihilating polynomials of $\mathcal{Y}_{W\dot{W}}$.

Show that the minimal annihilating polynomial for a genop $\mathcal{Y}_{W\dot{W}}$ is unique in the sense that if $p_1(s), p_2(s)$ are both minimal annihilating polynomials for $\mathcal{Y}_{W\dot{W}}$ then $p_2(s) = \lambda p_1(s)$, where $\lambda$ is nonzero.

**Answer**

If $p_1(s), p_2(s)$ are both minimal annihilating polynomials for $\mathcal{Y}_{W\dot{W}}$ and $p_2(s) \neq \lambda p_1(s)$, then $p_2(s) = p_1(s)q(s) + a(s)$, where $a(s)$ has degree lower than that of $p_1(s)$. By Theorem 65, we have $p_2(\mathcal{Y}_{W\dot{W}}) = p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}}) + \dot{a}(\mathcal{Y}_{W\dot{W}})$. Now $p_1(\mathcal{Y}_{W\dot{W}})$ is decoupled so that $p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}})$ is decoupled. Since $\mathcal{Y}_{W\dot{W}}$ is a genop and $p_2(\mathcal{Y}_{W\dot{W}})$ is also decoupled, we must have
$$p_1(\mathcal{Y}_{W\dot{W}}) = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W} = p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}}) = p_2(\mathcal{Y}_{W\dot{W}}).$$

Now by Lemma 66

$$p_2(\mathcal{Y}_{W\dot{W}}) \circ W = (p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}})) \circ W = a(\mathcal{Y}_{W\dot{W}}) \circ W,$$

$$p_2(\mathcal{Y}_{W\dot{W}}) \times \dot{W} = (p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}})) \times \dot{W} = a(\mathcal{Y}_{W\dot{W}}) \times \dot{W}.$$

Therefore, by part 1 of Theorem 17,

$$a(\mathcal{Y}_{W\dot{W}}) = p_2(\mathcal{Y}_{W\dot{W}}) + \stackrel{\cdot}{w}(-1)^{\stackrel{\cdot}{w}} p_1(\mathcal{Y}_{W\dot{W}}) * q(\mathcal{Y}_{W\dot{W}}) = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}, \text{ so that } a(\mathcal{Y}_{W\dot{W}}) \text{ is decoupled.}$$
$$\text{But } a(s) \text{ has lower degree than } p_1(s) \text{ which is a contradiction unless } a(s) \text{ is the zero polynomial.}$$

**Problem 75.** We have the following fundamental fact for operators in relation to their invariant subspaces.

‘Let $\mathcal{Y}_W$ be an invariant subspace for an operator $A$ and let $p(s)$ be an annihilating polynomial for $A$. Then $p(s)$ also annihilates $A/\mathcal{Y}_W$ and $A|_{\mathcal{Y}_W}$. Further, if $p_1(s), p_2(s)$, annihilate $A/\mathcal{Y}_W$, $A|_{\mathcal{Y}_W}$, respectively, then $p_1(s)p_2(s)$ annihilates $A$.’

Prove the following analogue for genops in relation to their invariant subspaces.

**Theorem 69.** Let $\mathcal{Y}_W$ be an invariant subspace for a genop $\mathcal{Y}_{W\dot{W}}$ and let $p(s)$ be an annihilating polynomial for $\mathcal{Y}_{W\dot{W}}$. Then $p(s)$ also annihilates $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$ and $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$. Further, if $p_1(s), p_2(s)$, annihilate $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$, $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$, respectively, then $p_1(s)p_2(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$.

Answer

We need the following lemmas for the proof of the result. The first is routine, following immediately from the definition of the '*' and '↔' operations.

**Lemma 70.** Let $\mathcal{Y}_{W\dot{W}}, \mathcal{Y}'_{W\dot{W}}$ be genauts. If $\mathcal{Y}_{W\dot{W}} \circ \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$, then $\mathcal{Y}_{W\dot{W}} * \mathcal{Y}'_{W\dot{W}} = \mathcal{Y}_{W\dot{W}} * (\mathcal{Y}'_{W\dot{W}} \cap \mathcal{Y}_W)$.

**Lemma 71.** Let $p(s)$ be a polynomial in $s$, and let $\mathcal{Y}_W$ be conditioned (controlled) invariant in genaut $\mathcal{Y}_{W\dot{W}}$. Then $\mathcal{Y}_W$ is conditioned (controlled) invariant in genaut $p(\mathcal{Y}_{W\dot{W}})$.

*Proof.* We will consider only the conditioned invariant case, the controlled invariant case following by duality.

Let $\mathcal{Y}_{W\dot{W}}^1, \mathcal{Y}_{W\dot{W}}^2$ be genauts and let $\mathcal{Y}_W$ be conditioned invariant in them. We will show that

1. $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2$,

2. $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}^1 + \dot{w} \mathcal{Y}_{W\dot{W}}^2$,

3. $\mathcal{Y}_W$ is conditioned invariant in $\lambda^{\dot{w}} \mathcal{Y}_{W\dot{W}}^1$,

It would follow that if $p(s)$ is a polynomial in $s$, then $\mathcal{Y}_W$ is conditioned invariant in $p(\mathcal{Y}_{W\dot{W}}^1)$.

1. We have $\mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2) = \mathcal{Y}_W \leftrightarrow [(\mathcal{Y}_{W\dot{W}}^1)_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}}] = [\mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1)_{WW_1}] \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}} \subseteq (\mathcal{Y}_W)_{W_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}} \subseteq (\mathcal{Y}_W)_{\dot{W}}$.

2. We have $\mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1 + \dot{w} \mathcal{Y}_{W\dot{W}}^2) \subseteq (\mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}}^1) + (\mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}}^2) \subseteq (\mathcal{Y}_W)_{\dot{W}}$.

3. When $\lambda \neq 0$, it is clear that $\mathcal{Y}_W \leftrightarrow \lambda^{\dot{w}} \mathcal{Y}_{W\dot{W}}^1 = \mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}}^1 \subseteq (\mathcal{Y}_W)_{\dot{W}}$.

Next, to consider the case where $\lambda = 0$, note that $\mathcal{Y}_W$ being conditioned invariant in $\mathcal{Y}_{W\dot{W}}^1$ implies that $(\mathcal{Y}_W)_{\dot{W}} \supseteq \mathcal{Y}_W \leftrightarrow \mathcal{Y}_{W\dot{W}}^1 \supseteq \mathcal{Y}_{W\dot{W}}^1 \times \dot{W}$. We know that $0^{\dot{W}} \mathcal{Y}_{W\dot{W}}^1 = \mathcal{Y}_{W\dot{W}}^1 \circ W \oplus \mathcal{Y}_{W\dot{W}}^1 \times \dot{W}$, so that $\mathcal{Y}_W \leftrightarrow 0^{\dot{W}} \mathcal{Y}_{W\dot{W}}^1 = \mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1 \circ W \oplus \mathcal{Y}_{W\dot{W}}^1 \times \dot{W}) = \mathcal{Y}_{W\dot{W}}^1 \times \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$.

□

**Lemma 72.** Let $p(s)$ be a polynomial in $s$, and let $\mathcal{Y}_{W\dot{W}}$ be a genaut. We then have the following.

1. If $\mathcal{Y}_W$ is conditioned invariant in $\mathcal{Y}_{W\dot{W}}$, then $p(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) = p(\mathcal{Y}_{W\dot{W}}) \cap \mathcal{Y}_W$.

2. If $\mathcal{Y}_W$ is controlled invariant in $\mathcal{Y}_{W\dot{W}}$, then $p(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) = p(\mathcal{Y}_{W\dot{W}}) + (\mathcal{Y}_W)_{\dot{W}}$.
*Proof.* We will consider only the conditioned invariant case, the controlled invariant case following by duality.
Let $\mathcal{Y}_{W\dot{W}}^1, \mathcal{Y}_{W\dot{W}}^2$ be genets and let $\mathcal{Y}_W$ be conditioned invariant in them. We will show that

$$1. (\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2) \cap \mathcal{Y}_W = (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W) * (\mathcal{Y}_{W\dot{W}}^2 \cap \mathcal{Y}_W).$$

$$2. (\mathcal{Y}_{W\dot{W}}^1 + \dot{\omega} \mathcal{Y}_{W\dot{W}}^2) \cap \mathcal{Y}_W = (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W) + \dot{\omega} (\mathcal{Y}_{W\dot{W}}^2 \cap \mathcal{Y}_W).$$

$$3. (\lambda^{\dot{w}} \mathcal{Y}_{W\dot{W}}^1) \cap \mathcal{Y}_W = \lambda^{\dot{w}} (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W).$$

It would follow that if $p(s)$ is a polynomial in $s$ then $p(\mathcal{Y}_{W\dot{W}}^1) \cap \mathcal{Y}_W = p(\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W)$.

$$1. \text{ We have } (\mathcal{Y}_{W\dot{W}}^1 * \mathcal{Y}_{W\dot{W}}^2) \cap \mathcal{Y}_W \equiv ((\mathcal{Y}_{W\dot{W}}^1)_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}}) \cap \mathcal{Y}_W \\ = (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}}. \text{ Now by conditioned invariance of } \mathcal{Y}_W \text{ in } \mathcal{Y}_{W\dot{W}}^1, \\ \mathcal{Y}_W \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1)_{WW_1} = ((\mathcal{Y}_{W\dot{W}}^1)_{WW_1} \cap \mathcal{Y}_W) \circ W_1 \subseteq (\mathcal{Y}_W)_{W_1}. \text{ So } (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W)_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}} \\ = (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W)_{WW_1} \leftrightarrow ((\mathcal{Y}_{W\dot{W}}^2)_{W_1\dot{W}} \cap \mathcal{Y}_{W_1})_{W_1\dot{W}} = (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W) * (\mathcal{Y}_{W\dot{W}}^2 \cap \mathcal{Y}_W).$$

$$2. \text{ (This does not require conditioned invariance of } \mathcal{Y}_W \text{.)}$$

That RHS is contained in LHS is immediate from the definition of the $+_{\dot{w}}$ operation.

Let $(f_W, g_{\dot{W}}) \in \text{LHS}$.

Then $f_W \in (\mathcal{Y}_{W\dot{W}}^1 \circ W) \cap (\mathcal{Y}_{W\dot{W}}^2 \circ W) \cap \mathcal{Y}_W$.

Further we must have, for some $g_{\dot{W}}^1, g_{\dot{W}}^2$, that $g_{\dot{W}}^1 + g_{\dot{W}}^2 = g_{\dot{W}}$, $(f_W, g_{\dot{W}}^1) \in \mathcal{Y}_{W\dot{W}}^1$, $(f_W, g_{\dot{W}}^2) \in \mathcal{Y}_{W\dot{W}}^2$. But then it is clear that $(f_W, g_{\dot{W}}^1) \in \mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W$, $(f_W, g_{\dot{W}}^2) \in \mathcal{Y}_{W\dot{W}}^2 \cap \mathcal{Y}_W$.

Thus $(f_W, g_{\dot{W}}) \in (\mathcal{Y}_{W\dot{W}}^1 \cap \mathcal{Y}_W) + \dot{\omega} (\mathcal{Y}_{W\dot{W}}^2 \cap \mathcal{Y}_W) = \text{RHS.}$

$$3. \text{ This is routine. It is clear that it works out right also for } \lambda = 0.$$

□

*Proof.* (of Theorem 69)

In the solution of Problem 64, we have seen that $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W = \mathcal{Y}_{W\dot{W}} \cap (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$ and $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W \oplus (\mathcal{Y}_W)_{\dot{W}})$, if $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}$.

By Lemma 71, $\mathcal{Y}_W$ is invariant in $p(\mathcal{Y}_{W\dot{W}})$ and by Lemma 72, $p(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W) = p(\mathcal{Y}_{W\dot{W}}) \cap \mathcal{Y}_W$ and $p(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) = p(\mathcal{Y}_{W\dot{W}}) + (\mathcal{Y}_W)_{\dot{W}}$. It follows that if $p(\mathcal{Y}_{W\dot{W}})$ is decoupled, so are $p(\mathcal{Y}_{W\dot{W}}) \cap \mathcal{Y}_W$ and $p(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}})$.

Suppose $p_1(s), p_2(s)$ annihilate $(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}})$, $(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W)$, respectively. Using the fact that if $\mathcal{Y}_{W\dot{W}} \subseteq \mathcal{Y}'_{W\dot{W}}$, then $p(\mathcal{Y}_{W\dot{W}}) \subseteq p(\mathcal{Y}'_{W\dot{W}})$, we will show that $p_1(\mathcal{Y}_{W\dot{W}}) \circ \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$, so that $p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}}) = p_1(\mathcal{Y}_{W\dot{W}}) * (p_2(\mathcal{Y}_{W\dot{W}}) \cap \mathcal{Y}_W)$ and thence that $p_1(s)p_2(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$.

We have $p_1(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}})$ decoupled so that $p_1(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) = p_1(\mathcal{Y}_{W\dot{W}}) + (\mathcal{Y}_W)_{\dot{W}}$ (using Lemma 72) $= p_1((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \circ W \oplus p_1((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \times \dot{W} = ((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \circ W \oplus ((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \times \dot{W}$ (using Lemma 66). Therefore $p_1(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \circ \dot{W} = ((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \times \dot{W} = ((\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \times \dot{W} + (\mathcal{Y}_W)_{\dot{W}} = (\mathcal{Y}_W)_{\dot{W}}$ (since $\mathcal{Y}_{W\dot{W}} \times \dot{W} \subseteq (\mathcal{Y}_W)_{\dot{W}}$ by the invariance of $\mathcal{Y}_W$ in $\mathcal{Y}_{W\dot{W}}$ and by using Theorem 1). It follows that $p_1(\mathcal{Y}_{W\dot{W}}) \circ \dot{W} \subseteq (p_1(\mathcal{Y}_{W\dot{W}}) + (\mathcal{Y}_W)_{\dot{W}}) \circ \dot{W} = p_1(\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}) \circ \dot{W} = (\mathcal{Y}_W)_{\dot{W}}$. Therefore $p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}}) = p_1(\mathcal{Y}_{W\dot{W}}) * (p_2(\mathcal{Y}_{W\dot{W}}) \cap \mathcal{Y}_W) = p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W)$ (using Lemma 70). But $p_2(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W)$ is decoupled so that $p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W)$ is decoupled (using part 1 of Problem 73). Thus $p_1(\mathcal{Y}_{W\dot{W}}) * p_2(\mathcal{Y}_{W\dot{W}})$ is decoupled and therefore $p_1(s)p_2(s)$ annihilates $\mathcal{Y}_{W\dot{W}}$.
### 15.5. Polynomials of $\mathcal{E}$-linked spaces

*(This subsection is based on Subsection 13.7 and the preceding subsections in this section)

We will now describe how operations on a GDS $\mathcal{V}_{W\dot{W}M}$ transfer across an $\mathcal{E}$-linkage to another GDS $\mathcal{V}_{P\dot{P}M}$ and, similarly, in the case of generalized autonomous systems $\mathcal{V}_{W\dot{W}}$, $\mathcal{V}_{P\dot{P}}$. In this subsection our primary concern is with polynomials of $\mathcal{E}$-linked genauts. The main result of this subsection, Theorem 74, states that polynomials, without constant terms, of $\mathcal{E}$-linked spaces continue to be $\mathcal{E}$-linked. Under extra conditions this is true for even the ones with constant terms. This implies that annihilation of a genaut by a polynomial, holds true across $\mathcal{E}$-linkages. So eigenvalue, eigenvector computations can be carried out through the most convenient emulators that we decide to choose. The qualification, about not having constant terms, implies that even if the original dynamical system has zero eigenvalues, a convenient emulator may not have (see Problem 92). (Emulators corresponding to state variables reflect the original dynamical system more accurately in this regard.) However, this is not a serious disadvantage of the general emulators. It only means that zero eigenvector computations which, in the case of large systems based on graphs, are topological and therefore very inexpensive, may have to be done directly on the original dynamical system (see Problem 93). We deal only with mutual emulators in this subsection. We say $\{\mathcal{V}_{W\dot{W}}, \mathcal{V}_{P\dot{P}}\}$ is $\mathcal{E}$-linked while talking of such emulators.

**Problem 76.** The $\mathcal{E}$-linkage is analogous to similarity transformation but much less restrictive in scope. For instance, it permits linking between systems with different numbers of variables (i.e., $|W|$ and $|P|$ need not be the same). If one thinks of $\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{W\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2$ as analogous to $T^{-1}AT$, $\mathcal{V}_{WP}^1$ takes on the role of $T^{-1}$, while $\mathcal{V}_{\dot{W}\dot{P}}^2$ that of $T$. Some flexibility is provided by $\mathcal{V}_{WP}^1, \mathcal{V}_{\dot{W}\dot{P}}^2$ to treat the $W, P$ variables differently from $\dot{W}, \dot{P}$ variables. This is appropriate because the derivatives of dynamic variables usually satisfy constraints of the dynamic variables and some additional ones. However, the cancellation that occurs when we multiply $T^{-1}AT$ by $T^{-1}AT$ resulting in $T^{-1}A^2T$ is mimicked by $\mathcal{E}$-linkage. Prove the following lemma and thence show that $\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{W\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2$ when 'multiplied' by itself results in $\mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{W\dot{W}}^{(2)} \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2$.

**Lemma 73.** Let $\mathcal{V}_{AB}, \mathcal{V}_{BC}$ be linkages and let linkages $\mathcal{V}_{BQ}^1, \mathcal{V}_{BQ}^2$ satisfy

$\mathcal{V}_{BQ}^1 \supseteq \mathcal{V}_{BQ}^2, \quad \mathcal{V}_{BQ}^2 \circ B \supseteq \mathcal{V}_{AB} \circ B, \quad \mathcal{V}_{BQ}^1 \times B \subseteq \mathcal{V}_{BC} \times B, A, B, C, Q, \text{ being pairwise disjoint. Then} \\
\mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BC} = \mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BQ}^2 \leftrightarrow (\mathcal{V}_{BQ}^1)_{B_1Q} \leftrightarrow (\mathcal{V}_{BC})_{B_1C}.$

**Answer**

*Proof.* Let $(f_A, h_C) \in \mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BC}$. Then there exists $g_B$ such that $(f_A, g_B) \in \mathcal{V}_{AB}$, $(g_B, h_C) \in \mathcal{V}_{BC}$. Since $\mathcal{V}_{BQ}^2 \circ B \supseteq \mathcal{V}_{AB} \circ B$, there exists $k_Q$ such that $(g_B, k_Q) \in \mathcal{V}_{BQ}^2$ and since $\mathcal{V}_{BQ}^1 \supseteq \mathcal{V}_{BQ}^2$, it follows that $(g_B, k_Q) \in \mathcal{V}_{BQ}^1$. Noting that $B_1$ is a copy of $B$, we must have $(g_{B_1}, k_Q) \in (\mathcal{V}_{BQ}^1)_{B_1Q}$. Thus we have $(f_A, g_B) \in \mathcal{V}_{AB}$, $(g_B, k_Q) \in \mathcal{V}_{BQ}^2$, $(g_{B_1}, k_Q) \in (\mathcal{V}_{BQ}^1)_{B_1Q}$, $(g_{B_1}, h_C) \in (\mathcal{V}_{BC})_{B_1C}$, so that $(f_A, h_C) \in \mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BQ}^2 \leftrightarrow (\mathcal{V}_{BQ}^1)_{B_1Q} \leftrightarrow (\mathcal{V}_{BC})_{B_1C}$. Thus $\mathrm{LHS} \subseteq \mathrm{RHS}$.

Next, let $(f_A, h_C) \in \mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BQ}^2 \leftrightarrow (\mathcal{V}_{BQ}^1)_{B_1Q} \leftrightarrow (\mathcal{V}_{BC})_{B_1C}$. Then there exist $(f_A, g_B) \in \mathcal{V}_{AB}$, $(g_B, k_Q) \in \mathcal{V}_{BQ}^2$, $(g'_{B_1}, k_Q) \in (\mathcal{V}_{BQ}^1)_{B_1Q}$, $(g'_{B_1}, h_C) \in (\mathcal{V}_{BC})_{B_1C}$. But this means $(g_{B_1}, k_Q), (g'_{B_1}, k_Q) \in (\mathcal{V}_{BQ}^1)_{B_1Q}$ (since $\mathcal{V}_{BQ}^2 \subseteq \mathcal{V}_{BQ}^1$). Therefore, $(g_{B_1} - g'_{B_1}) \in (\mathcal{V}_{BQ}^1)_{B_1Q} \times B_1 \subseteq (\mathcal{V}_{BC})_{B_1C} \times B_1$. We conclude that $(g_{B_1}, h_C) \in (\mathcal{V}_{BC})_{B_1C}$ and equivalently $(g_B, h_C) \in \mathcal{V}_{BC}$. Thus $(f_A, g_B) \in \mathcal{V}_{AB}$, $(g_B, h_C) \in (\mathcal{V}_{BC})$. So $(f_A, h_C) \in \mathcal{V}_{AB} \leftrightarrow \mathcal{V}_{BC}$. Thus $\mathrm{RHS} \subseteq \mathrm{LHS}$.

□

We have,

$$ \mathcal{V}_{WP}^1 \leftrightarrow \mathcal{V}_{W\dot{W}}^{(2)} \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2 = \mathcal{V}_{WP}^1 \leftrightarrow (\mathcal{V}_{W\dot{W}} * \mathcal{V}_{W\dot{W}}) \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2 = \mathcal{V}_{WP}^1 \leftrightarrow (\mathcal{V}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{V}_{W\dot{W}})_{W_1\dot{W}} \leftrightarrow \mathcal{V}_{\dot{W}\dot{P}}^2. $$

Using properties of $\mathcal{E}$-linkage, we note that

$$ (\mathcal{V}_{WP}^1)_{W_1P_1} \supseteq (\mathcal{V}_{\dot{W}\dot{P}}^2)_{W_1P_1}, (\mathcal{V}_{\dot{W}\dot{P}}^2)_{W_1P_1} \circ W_1 \supseteq (\mathcal{V}_{W\dot{W}})_{WW_1} \circ W_1, (\mathcal{V}_{WP}^1)_{W_1P_1} \times W_1 \subseteq (\mathcal{V}_{W\dot{W}})_{W_1\dot{W}} \times W_1. $$

Now we expand $(\mathcal{V}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{V}_{W\dot{W}})_{W_1\dot{W}}$, using Lemma 73 as
$$ (\mathcal{Y}_{W\dot{W}})_{WW_1} \leftrightarrow (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{W_1P_1} \leftrightarrow (\mathcal{Y}_{WP}^1)_{W_1P_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1\dot{W}}. $$

So $\mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^2 * \mathcal{Y}_{W\dot{W}}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2$ can be rewritten as

$$ (\mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}))_{WW_1} \leftrightarrow (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{W_1P_1} \leftrightarrow ((\mathcal{Y}_{WP}^1)_{W_1P_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2), $$

i.e., as $(\mathcal{Y}_{P\dot{P}})_{PP_1} \leftrightarrow (\mathcal{Y}_{P\dot{P}})_{P_1\dot{P}} = \mathcal{Y}_{P\dot{P}} * \mathcal{Y}_{P\dot{P}} = \mathcal{Y}_{P\dot{P}}^{(2)}$.

**Problem 77.** Prove the following result which states, among other things, that a polynomial without constant term annihilates a genop iff it annihilates a genop $\mathcal{E}$-linked to it.

**Theorem 74.** Let $p(s)$ be a polynomial with no constant term. Let $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_{P\dot{P}}$ be genops such that $\{\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}}\}$ is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. We then have the following.

1. $\{p(\mathcal{Y}_{W\dot{W}}), p(\mathcal{Y}_{P\dot{P}})\}$ is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$ and therefore, if one of $p(\mathcal{Y}_{W\dot{W}}), p(\mathcal{Y}_{P\dot{P}})$ is decoupled so would the other be.

2. In general $\mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)}$ would not be $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. If however,

$$ \begin{aligned} \mathcal{Y}_{W\dot{W}} \circ W &= (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W, \quad \mathcal{Y}_{W\dot{W}} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W, \quad \mathcal{Y}_{P\dot{P}} \circ P = (\mathcal{Y}_{P\dot{P}} \circ \dot{P})_P, \quad \mathcal{Y}_{P\dot{P}} \times P = (\mathcal{Y}_{P\dot{P}} \times \dot{P})_P, \\ \mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)} &\text{ would be } \mathcal{E}\text{-linked through } \mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2 \text{ and } \{p(\mathcal{Y}_{W\dot{W}}), p(\mathcal{Y}_{P\dot{P}})\} \text{ would be } \mathcal{E}\text{-linked} \\ &\text{through } \mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2, \text{ even if } p(s) \text{ has a constant term. If one of } p(\mathcal{Y}_{W\dot{W}}), p(\mathcal{Y}_{P\dot{P}}) \text{ is decoupled so would} \\ &\text{the other be.} \end{aligned} $$

Answer

We note that $\mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 = (\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{W\dot{W}}^1) \oplus (\mathcal{Y}_{\dot{W}\dot{P}}^2 \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2)$, so that if $\{\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}}\}$ are $\mathcal{E}$-linked and one of them is decoupled so would the other be.

As is the case of results involving general polynomials (see proof of Lemma 66, Problems 70, 71), we examine what happens in the case of $\mathcal{Y}_{W\dot{W}}^{(k)}, \lambda_j^w \mathcal{Y}_{W\dot{W}}^{(j)}, \lambda_j^w \mathcal{Y}_{W\dot{W}}^{(j)} * \lambda_k^w \mathcal{Y}_{W\dot{W}}^{(k)}, \lambda_j^w \mathcal{Y}_{W\dot{W}}^{(j)} + \dot{\lambda}_k^w \mathcal{Y}_{W\dot{W}}^{(k)}$. We do this in the following lemmas from which Theorem 74 is immediate.

**Lemma 75.** Let $\{\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}}\}$ be $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. Then

1. $\{\lambda^w(\mathcal{Y}_{W\dot{W}}), \lambda^p(\mathcal{Y}_{P\dot{P}})\}$ is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$.

2. $\{\mathcal{Y}_{W\dot{W}}^{(k)}, (\mathcal{Y}_{P\dot{P}})^{(k)}\}, k \neq 0$ is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$.

3. If further,

$$ \begin{aligned} \mathcal{Y}_{W\dot{W}} \circ W &= (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W, \quad \mathcal{Y}_{W\dot{W}} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W, \quad \mathcal{Y}_{P\dot{P}} \circ P = (\mathcal{Y}_{P\dot{P}} \circ \dot{P})_P, \quad \mathcal{Y}_{P\dot{P}} \times P = (\mathcal{Y}_{P\dot{P}} \times \dot{P})_P, \\ \text{then we have } \{\mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)}\} \mathcal{E}\text{-linked through } \mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2. \end{aligned} $$

*Proof.* 1. When $\lambda = 0$, $\lambda^w(\mathcal{Y}_{W\dot{W}}) = \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}$.

$$ \begin{aligned} & \text{From the definition of } \mathcal{E}\text{-linkage, it follows that } \lambda^w(\mathcal{Y}_{W\dot{W}}) \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \\ &= (\mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}) \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) = \mathcal{Y}_{P\dot{P}} \circ P \oplus \mathcal{Y}_{P\dot{P}} \times \dot{P} = \lambda^p(\mathcal{Y}_{P\dot{P}}). \end{aligned} $$

Next, let $\lambda \neq 0$ and let $(f_P, \lambda g_{\dot{P}}) \in \lambda^w(\mathcal{Y}_{W\dot{W}}) \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2)$.

Then there exist

$$ (h_W, \lambda k_{\dot{W}}) \in \lambda^w(\mathcal{Y}_{W\dot{W}}), \quad (h_W, f_P) \in \mathcal{Y}_{WP}^1, \quad (\lambda k_{\dot{W}}, \lambda g_{\dot{P}}) \in \mathcal{Y}_{\dot{W}\dot{P}}^2. $$

Therefore,

$$ (h_W, k_{\dot{W}}) \in \mathcal{Y}_{W\dot{W}}, \quad (h_W, f_P) \in \mathcal{Y}_{WP}^1, \quad (k_{\dot{W}}, g_{\dot{P}}) \in \mathcal{Y}_{\dot{W}\dot{P}}^2, $$
so that $(f_P, g_P) \in \mathcal{Y}_{P\dot{P}}$. It follows that

$$ (f_P, \lambda g_P) \in \lambda \dot{P}(\mathcal{Y}_{P\dot{P}}), \text{ i.e., } \lambda \dot{w}(\mathcal{Y}_{W\dot{W}}) \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) \subseteq \lambda \dot{P}(\mathcal{Y}_{P\dot{P}}). $$

The same argument, using $\mathcal{Y}_{P\dot{P}} \leftrightarrow (\mathcal{Y}_{WP}^1 \oplus \mathcal{Y}_{\dot{W}\dot{P}}^2) = \mathcal{Y}_{W\dot{W}}$, proves the reverse containment.

2. We are given that $\{\mathcal{Y}_{W\dot{W}}, \mathcal{Y}_{P\dot{P}}\}$, is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. Suppose $\{\mathcal{Y}_{W\dot{W}}^{(k)}, (\mathcal{Y}_{P\dot{P}}^{(k)})\}, 0 < k < r$, is $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. We know that the statement is true for $k=1$. We will show that the statement is true also for $k=r$.

By Lemma 66, we have that for any genaut $\mathcal{Y}_{W\dot{W}}$,

$$ \mathcal{Y}_{W\dot{W}}^{(k)} \circ \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W}, \quad \mathcal{Y}_{W\dot{W}}^{(k)} \times W \supseteq \mathcal{Y}_{W\dot{W}} \times W. $$

We therefore have

$$ \mathcal{Y}_{WP}^1 \times W \subseteq \mathcal{Y}_{W\dot{W}} \times W \subseteq \mathcal{Y}_{W\dot{W}}^{(k)} \times W \text{ and } \mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W} \supseteq \mathcal{Y}_{W\dot{W}} \circ \dot{W} \supseteq \mathcal{Y}_{W\dot{W}}^{(k)} \circ \dot{W}. $$

We have,

$$ \mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{W\dot{W}}^{(r)} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 = \mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^{(r-1)} * \mathcal{Y}_{W\dot{W}}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 = \mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^{(r-1)})_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2. $$

Now we expand $(\mathcal{Y}_{W\dot{W}}^{(r-1)})_{WW_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1\dot{W}}$, using Lemma 73 as

$$ (\mathcal{Y}_{W\dot{W}}^{(r-1)})_{WW_1} \leftrightarrow (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{W_1P_1} \leftrightarrow (\mathcal{Y}_{WP}^1)_{W_1'P_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1'\dot{W}}. $$

So $\mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^{(r-1)} * \mathcal{Y}_{W\dot{W}}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2$ can be rewritten as

$$ (\mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}_{W\dot{W}}^{(r-1)})_{WW_1} \leftrightarrow (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{W_1P_1}) \leftrightarrow ((\mathcal{Y}_{WP}^1)_{W_1'P_1} \leftrightarrow (\mathcal{Y}_{W\dot{W}})_{W_1'\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2), $$

i.e., as

$$ ((\mathcal{Y}_{P\dot{P}}^{(r-1)})_{PP_1} \leftrightarrow (\mathcal{Y}_{P\dot{P}})_{P_1\dot{P}} = \mathcal{Y}_{P\dot{P}}^{(r-1)} * \mathcal{Y}_{P\dot{P}} = \mathcal{Y}_{P\dot{P}}^{(r)}). $$

3. We have $\mathcal{Y}_{W\dot{W}}^{(0)} := I_{W\dot{W}} + \mathcal{Y}_{W\dot{W}} \circ W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}$, so that

$$ \begin{aligned} \mathcal{Y}_{W\dot{W}}^{(0)} &:= \{(f_W, f_{\dot{W}}), f_W \in \mathcal{Y}_{W\dot{W}} \circ W\} + (\mathbf{0}_W \oplus \mathcal{Y}_{W\dot{W}} \times \dot{W}) \\ &= \{(f_W, f_{\dot{W}} + f'_{\dot{W}}), f_W \in \mathcal{Y}_{W\dot{W}} \circ W, f'_{\dot{W}} \in \mathcal{Y}_{W\dot{W}} \times \dot{W}\}. \end{aligned} $$

In order for $\{\mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)}\}$ to be $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$, we need the conditions

$\mathcal{Y}_{WP}^1 \circ W \supseteq \mathcal{Y}_{W\dot{W}}^{(0)} \circ W, \mathcal{Y}_{WP}^1 \times W \subseteq \mathcal{Y}_{W\dot{W}}^{(0)} \times W, \mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W} \supseteq \mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W}, \mathcal{Y}_{\dot{W}\dot{P}}^2 \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}}^{(0)} \times \dot{W}$, to hold when the corresponding conditions replacing $\mathcal{Y}_{W\dot{W}}^{(0)}$ by $\mathcal{Y}_{W\dot{W}}$, hold. But when $\mathcal{Y}_{W\dot{W}}$ is a genop, in general $\mathcal{Y}_{W\dot{W}}^{(0)} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W \neq \mathcal{Y}_{W\dot{W}} \times W$ and $\mathcal{Y}_{W\dot{W}}^{(0)} \circ \dot{W} = (\mathcal{Y}_{W\dot{W}} \circ W)_{\dot{W}} \neq \mathcal{Y}_{W\dot{W}} \circ \dot{W}$. So $\{\mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)}\}$ may not be $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$. We give an example to show that in general $\{\mathcal{Y}_{W\dot{W}}^{(0)}, \mathcal{Y}_{P\dot{P}}^{(0)}\}$ would not be $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$, when $\mathcal{Y}_{W\dot{W}}$, $\mathcal{Y}_{P\dot{P}}$ are genops. Suppose $h_W, g_P$ are such that $h_W \in \mathcal{Y}_{W\dot{W}} \circ W - (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W$, $(h_W, g_P) \in (\mathcal{Y}_{\dot{W}\dot{P}}^2)_{WP}, g_P \notin \mathcal{Y}_{P\dot{P}} \circ P$. It is clear that $g_P \in ((\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}_{W\dot{W}}^{(0)} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2) \circ \dot{P})_P$ but, $g_P \notin (\mathcal{Y}_{P\dot{P}}^{(0)} \circ \dot{P})_P$.

On the other hand, it is easy to see that when

$$ \mathcal{Y}_{W\dot{W}} \circ W = (\mathcal{Y}_{W\dot{W}} \circ \dot{W})_W, \quad \mathcal{Y}_{W\dot{W}} \times W = (\mathcal{Y}_{W\dot{W}} \times \dot{W})_W, \quad \mathcal{Y}_{P\dot{P}} \circ P = (\mathcal{Y}_{P\dot{P}} \circ \dot{P})_P, \quad \mathcal{Y}_{P\dot{P}} \times P = (\mathcal{Y}_{P\dot{P}} \times \dot{P})_P, $$
we have $\mathcal{Y}_{WP}^1 \leftrightarrow \{(f_W, f_W^\cdot + f_W^\prime), f_W \in \mathcal{Y}_{W\dot{W}} \circ W, f_W^\prime \in \mathcal{Y}_{W\dot{W}} \times \dot{W}\} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2$
$= \{(h_P, h_P^\cdot + h_P^\prime), h_P \in \mathcal{Y}_{P\dot{P}} \circ P, h_P^\prime \in \mathcal{Y}_{P\dot{P}} \times \dot{P}\}$.

□

**Lemma 76.** Let {$\mathcal{Y}'_{W\dot{W}}, \mathcal{Y}'_{P\dot{P}}$}, {$\tilde{\mathcal{Y}}_{W\dot{W}}, \tilde{\mathcal{Y}}_{P\dot{P}}$} be both $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$.

Then so is {$\mathcal{Y}'_{W\dot{W}} + \dot{w}\tilde{\mathcal{Y}}_{W\dot{W}}, \mathcal{Y}'_{P\dot{P}} + \dot{p}\tilde{\mathcal{Y}}_{P\dot{P}}$} $\mathcal{E}$-linked through $\mathcal{Y}_{WP}^1, \mathcal{Y}_{\dot{W}\dot{P}}^2$.

*Proof.* The result follows by using the distributivity of '$\leftrightarrow$' over the intersection-sum operation for appropriate conditions (Theorem 18). We have

$$ \begin{align*} \mathcal{Y}_{WP}^1 &\leftrightarrow (\mathcal{Y}'_{W\dot{W}} + \dot{w}\tilde{\mathcal{Y}}_{W\dot{W}}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 \\ &= ((\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}'_{W\dot{W}}) + \dot{w}(\mathcal{Y}_{WP}^1 \leftrightarrow \tilde{\mathcal{Y}}_{W\dot{W}})) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 \\ \text{(since } \mathcal{Y}_{WP}^1 \times W \subseteq \mathcal{Y}'_{W\dot{W}} \times W, \mathcal{Y}_{WP}^1 \times W \subseteq \tilde{\mathcal{Y}}_{W\dot{W}} \times W) \\ &= (\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}'_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2) + \dot{p}(\mathcal{Y}_{WP}^1 \leftrightarrow \tilde{\mathcal{Y}}_{W\dot{W}} \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2) \\ \text{(since } (\mathcal{Y}_{WP}^1 \leftrightarrow \mathcal{Y}'_{W\dot{W}}) \circ \dot{W} \subseteq \mathcal{Y}'_{W\dot{W}} \circ \dot{W} \subseteq \mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W}, (\mathcal{Y}_{WP}^1 \leftrightarrow \tilde{\mathcal{Y}}_{W\dot{W}}) \circ \dot{W} \subseteq \tilde{\mathcal{Y}}_{W\dot{W}} \circ \dot{W} \subseteq \mathcal{Y}_{\dot{W}\dot{P}}^2 \circ \dot{W}) \\ &= \mathcal{Y}'_{P\dot{P}} + \dot{p}\tilde{\mathcal{Y}}_{P\dot{P}}. \end{align*} $$

The same argument, because of the properties of $\mathcal{E}$-linkage being symmetric with respect to $W, P$ and $\dot{W}, \dot{P}$, works for showing

$$ \mathcal{Y}_{WP}^1 \leftrightarrow (\mathcal{Y}'_{P\dot{P}} + \dot{p}\tilde{\mathcal{Y}}_{P\dot{P}}) \leftrightarrow \mathcal{Y}_{\dot{W}\dot{P}}^2 = \mathcal{Y}'_{W\dot{W}} + \dot{w}\tilde{\mathcal{Y}}_{W\dot{W}}. $$

□

**16. Pole placement through state feedback and output injection**

*(This section is based on Section12, Subsection13.2, Section 14, Section 15.)*

A fundamental problem in control theory is to 'place poles' of a system in a desired region of the complex plane. The classical result here is that if the system is fully controllable, the poles can be placed where we wish using state feedback and dually, if the system is fully observable, the poles can be placed where we wish using output injection. There are neat refinements of this result when the system is not fully controllable or fully observable. In this section we derive implicit versions of this result. In Section 17, we address, in a limited way, the output feedback problem, where we have to shift poles to desired locations using only the input and output.

**Definition 77.** A dynamical system $\mathcal{Y}_{W\dot{W}UY}$ is regular (RDS) iff

$\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}$ is a USG, $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}$ is an LSG and $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W\dot{W} = \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times W\dot{W}$ is a genop.

It is easily verified that $\mathcal{Y}_{W\dot{W}UY}$ is regular iff its adjoint is regular. Further, if $\mathcal{Y}_{W\dot{W}UY}$ is of the state-input-output form defined by $\dot{w}=Aw+Bu, y=Cw$, it is regular since $\mathcal{Y}_{W\dot{W}UY} \circ W = \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W =$ $\mathcal{F}_W, \mathcal{Y}_{W\dot{W}UY} \times \dot{W} = \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} = \mathbf{0}_{\dot{W}}$ and $\mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W\dot{W}$ is defined by $\dot{w}=Aw$. Since $\mathcal{Y}_{W\dot{W}UY}$ being regular implies $\mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}, \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}$, being respectively a USG and an LSG, ideas from Section 14.2 are applicable. The first basic question is "Given an RDS $\mathcal{Y}_{W\dot{W}UY}$ what are
the genops which can be obtained by $(W,U)$- feedback and $(\dot{W},Y)$- injection?" This question has been
answered for general autonomous spaces in Problems 48 and 49 and therefore also for genops. In brief the
answer is:

The genop $\mathcal{Y}_{W\dot{W}}$ can be obtained by $(W,U)$- feedback iff it satisfies

$$
\mathcal{Y}_{W\dot{W}UY} \circ W \dot{W} \supseteq \mathcal{Y}_{W\dot{W}}; \quad \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}U \times \dot{W} \subseteq \mathcal{Y}_{W\dot{W}} \times \dot{W}.
$$

The genop $\mathcal{Y}_{W\dot{W}}$ can be obtained by $(\dot{W}, Y)$- injection iff it satisfies

$$
\mathcal{Y}_{W\dot{W}UY} \times W \dot{W} \subseteq \mathcal{Y}_{W\dot{W}}; \quad \mathcal{Y}_{W\dot{W}UY} \times W\dot{W}Y \circ W \supseteq \mathcal{Y}_{W\dot{W}} \circ W.
$$

In this section we address the question: "What polynomials can be made into minimal annihilating polynomials for genops obtainable by (W,U)- feedback and for those obtainable by (\dot{W},Y)- injection from the RDS $\mathcal{Y}_{W\dot{W}UY}$?"

**Problem 78.** When $\mathcal{Y}_{W\dot{W}UY}$ is of the state-input-output form defined by $\dot{w}=Aw+Bu, y=Cw$, and $\mathcal{Y}_W$ is its controllable subspace, we know that the minimal annihilating polynomial of $A/\mathcal{Y}_W$ has to be a factor of every annihilating polynomial of $\mathcal{Y}_{W\dot{W}}$ obtained by state feedback from $\mathcal{Y}_{W\dot{W}UY}$. We translate this idea to our notation below.

Let $\mathcal{Y}_{W\dot{W}UY}$ be an RDS and let $\mathcal{Y}_{W\dot{W}}^u \equiv \mathcal{Y}_{W\dot{W}UY} \circ W \dot{W}$. Let $\mathcal{Y}_W$ be the minimal invariant space of
$\mathcal{Y}_{W\dot{W}}^u$ containing $(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$. Let $\mathcal{Y}_{W\dot{W}} := (\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_{WU}) \circ W \dot{W}$ be a genop. with $\mathcal{Y}_{WU}$ such that
$\mathcal{Y}_{WU} \circ W = \mathcal{Y}_{W\dot{W}}^u \circ W$ (equivalently, let $\mathcal{Y}_{W\dot{W}}$ be a genop such that $\mathcal{Y}_{W\dot{W}} \subseteq \mathcal{Y}_{W\dot{W}}^u$, $\mathcal{Y}_{W\dot{W}} \circ W = \mathcal{Y}_{W\dot{W}}^u \circ W$).
Show that every annihilating polynomial of $\mathcal{Y}_{W\dot{W}}$ contains as a factor, the minimal annihilating polynomial
$p_u(s)$ of the genop $\mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$.

Answer

We have $\mathcal{Y}_{W\dot{W}} := (\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_{WU}) \circ W\dot{W}$ a genop, $\mathcal{Y}_{W\dot{W}}^u \circ W = (\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_{WU}) \circ W = \mathcal{Y}_{W\dot{W}}^u \circ W$ and
$\mathcal{Y}_W$ invariant in $\mathcal{Y}_{W\dot{W}}^u$. Therefore by Lemma 56, we have that $\mathcal{Y}_W$ is invariant in $\mathcal{Y}_{W\dot{W}}$, the spaces $\mathcal{Y}_{W\dot{W}} \cap$
$\mathcal{Y}_W, \mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}}$ are genops and $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}^u + (\mathcal{Y}_W)_{\dot{W}}$. By Theorem 69, every annihilating
polynomial of $\mathcal{Y}_{W\dot{W}}$ contains as a factor the minimal annihilating polynomial of $\mathcal{Y}_{W\dot{W}} + (\mathcal{Y}_W)_{\dot{W}} = \mathcal{Y}_{W\dot{W}}^u +$
$(\mathcal{Y}_W)_{\dot{W}}$.

**Problem 79.** When $\mathcal{Y}_{W\dot{W}UY}$ is of the state-input-output form defined by $\dot{w}=Aw+Bu, y=Cw$, and $\mathcal{Y}_W$ is its controllable subspace we know that through some state feedback $u=Fw$, with $F$ of column rank 1, we can make $(A+BF)|_{\mathcal{Y}_W}$ have any desired minimal annihilating polynomial $p_2(s)$ of degree equal to the dimension of $\mathcal{Y}_W$. Give a sketch of this procedure in a form which can be naturally translated to our framework.

Answer

The main object which we use for modifying matrices is a 'full Krylov sequence'.

**Definition 78.** A sequence $\omega = (w^0, \dots, w^k)$ of $k \times 1$ vectors is said to be a full Krylov sequence for a $k \times k$ matrix $A$, iff $w^j = A^j w^0, j = 1, \dots, k$, and $w^0, \dots, w^{k-1}$, are linearly independent.

Every matrix does not have a Krylov sequence. However, because we restrict our transformations of the
matrices in question to the controllable space, we can start with a matrix which has a full Krylov sequence
and continue in our algorithm with only such matrices. When a matrix has a full Krylov sequence, it is fully
determined by the latter.

Let us consider a system described by the state-input-output equations. The classical state feedback of
[22, 23] can be thought of as a modification of a full Krylov sequence $(w^0, \dots, w^k)$ to another $(v^0, \dots, v^k)$,
through the rule

$$
(w^0 | \cdots | w^k) \begin{bmatrix} \lambda_0 & \lambda_1 & \cdots & \cdots & \lambda_k \\ 0 & \lambda_0 & \cdots & \cdots & \lambda_{k-1} \\ \vdots & \vdots & \ddots & \ddots & \vdots \\ 0 & 0 & \cdots & 0 & \lambda_0 \end{bmatrix} = (v^0 | \cdots | v^k), \lambda_0 \neq 0. \quad (56)
$$
If $s^k + d_1s^{k-1} + \cdots + d_k$, $s^k + b_1s^{k-1} + \cdots + b_k$, are respectively the annihilating polynomials of the matrices with full Krylov sequences $(w^0, \cdots, w^k)$, $(v^0, \cdots, v^k)$, then we must have

$$ \begin{bmatrix} \lambda_0 & \lambda_1 & \cdots & \cdots & \lambda_k \\ 0 & \lambda_0 & \cdots & \cdots & \lambda_{k-1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & \lambda_0 \end{bmatrix} \begin{pmatrix} b_k \\ b_{k-1} \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} d_k \\ d_{k-1} \\ \vdots \\ 1 \end{pmatrix}. \quad (57) $$

Further, one can show that the condition in Equation 57 is actually equivalent to the one in Equation 56. The above corresponds to the matrix $A$ being transformed through state feedback to $A + BF$, where the state feedback matrix $F$ has rank one.

Coefficient matrix of the kind in Equation 57 is a triangular Toeplitz matrix and is entirely determined by the sequence $(1, \lambda_1, \cdots, \lambda_k)$. It is conveniently manipulated treating it as though it is the polynomial $s^k + \lambda_1 s^{k-1} + \cdots + \lambda_k$. Multiplication of such matrices corresponds to multiplying the corresponding polynomials modulo $s^{k+1}$. One of the useful consequences is that Equation 57 can be written as

$$ \begin{bmatrix} 1 & b_1 & \cdots & \cdots & b_k \\ 0 & 1 & \cdots & \cdots & b_{k-1} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & 1 \end{bmatrix} \begin{pmatrix} \lambda_k \\ \lambda_{k-1} \\ \vdots \\ 1 \end{pmatrix} = \begin{pmatrix} d_k \\ d_{k-1} \\ \vdots \\ 1 \end{pmatrix}. \quad (58) $$

A solution to this equation can be used to build a sequence $(1, \lambda_1, \cdots, \lambda_k)$ and transform the full Krylov sequence $(w^0, \cdots, w^k)$ to $(v^0, \cdots, v^k)$. The latter will have $s^k + b_1s^{k-1} + \cdots + b_k$ as the minimal annihilating polynomial.

**Problem 80.** We translate the sketch of Problem 79 to our notation below.

Let $\mathcal{Y}_{W\dot{W}UY}$ be an RDS and let $\mathcal{Y}_{W\dot{W}}^u \equiv \mathcal{Y}_{W\dot{W}UY} \circ W\dot{W}$. Let $\mathcal{Y}_W$ be the minimal invariant space of $\mathcal{Y}_{W\dot{W}}^u$ containing $(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$ and let $\mathcal{Y}_W^{com} \equiv \mathcal{Y}_{W\dot{W}}^u \times W \cap (\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$. Let $p_2(s)$ be any polynomial of degree equal to $r(\mathcal{Y}_W) - r(\mathcal{Y}_W^{com})$. Show that $p_2(s)$ can be made into the minimal annihilating polynomial of a genop $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$, where $\mathcal{Y}_{W\dot{W}} \subseteq \mathcal{Y}_{W\dot{W}}^u$.

Answer

We give a solution in our framework through the following subsections.

**16.1. Constructing a genop $\mathcal{Y}_{W\dot{W}} \cap \mathcal{Y}_W$ with desired annihilating polynomial**

Our construction involves the idea of a 'basic sequence' which is analogous to the full Krylov sequence for an operator $A$. We will begin by building a genop $\mathcal{Y}_{W\dot{W}}^{start}$ (Algorithm V) with the annihilating polynomial of degree equal to $r(\mathcal{Y}_W) - r(\mathcal{Y}_W^{com})$, where $\mathcal{Y}_W^{com} \equiv \mathcal{Y}_{W\dot{W}}^u \times W \cap (\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$. Next, we modify the basic sequence of $\mathcal{Y}_{W\dot{W}}^{start}$ suitably (Equation 59 below) to obtain another sequence whose genop $\mathcal{Y}_{W\dot{W}}^{end}$ has the desired minimal annihilating polynomial.

**Definition 79.** Let $\mathcal{Y}_{W\dot{W}}^u$ be a USG and let $\mathcal{Y}_W$ be invariant in it. We say $w_W^0, w_W^1, \cdots, w_W^k$, is a basic sequence in $\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_W$, iff

$w_W^0, w_W^1, \cdots, w_W^{k-1}$, are linearly independent,

$$ w_W^j \in \mathcal{Y}_W, \quad j = 0, 1, \cdots, k $$

$$ w_{\dot{W}}^{j+1} := (w_W^{j+1})_{\dot{W}} \in (w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u), \quad j = 0, 1, \cdots, k-1 \quad \text{and} \quad w_W^k \in \mathcal{Y}_W. $$

$$ \mathcal{Y}_W = \text{sp}\{w_W^j, j = 0, 1, 2, \cdots, k-1\} + \mathcal{Y}_W^{com}, \text{ where } \mathcal{Y}_W^{com} := \mathcal{Y}_{W\dot{W}}^u \times W \cap (\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W $$

We now construct a basic sequence in $\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_W$, and from it build a genop $\mathcal{Y}_{W\dot{W}}^{start}$, for which it is a basic sequence.
### 16.1.1. Building the genop $\mathcal{Y}_{W\dot{W}}^{start}$

#### Algorithm V

#### Algorithm Basic sequence

Input: genaut $\mathcal{Y}_{W\dot{W}}^u$, that is USG but not LSG.
Output:

1. $\mathcal{Y}_{W}$, the minimal space invariant in $\mathcal{Y}_{W\dot{W}}^u$, containing $(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$,

2. A basic sequence $w_W^0, w_W^1, \dots, w_W^k$, in $\mathcal{Y}_{W\dot{W}}^u \cap \mathcal{Y}_{W}$,

3. The genop $\mathcal{Y}_{W\dot{W}}^{start}$.

Step 1. Pick $w_W^0 \in [(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W - \mathcal{Y}_{W\dot{W}}^u \times W]$.

Step 2. Pick, if possible $w_{\dot{W}}^{j+1} \in (w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u)$, and $w_W^0, w_W^1, \dots, w_W^{j+1}$ independent modulo $\mathcal{Y}_{W}^{com}$.

Stop if $w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u \subseteq (\text{sp}\{w_W^i, i = 0, 1, 2, \dots, j\} + \mathcal{Y}_{W}^{com})_{\dot{W}}$.

$k \leftarrow j + 1$.

$\mathcal{Y}_{W} \leftarrow \text{sp}\{w_W^j, j = 0, 1, 2, \dots, k-1\} + \mathcal{Y}_{W}^{com}$.

$\mathcal{Y}_{W\dot{W}}^{start} \leftarrow \text{sp}\{(w_W^j, w_{\dot{W}}^{j+1}), j = 0, 1, 2, \dots, k-1\} + (\mathcal{Y}_{W}^{com} \oplus (\mathcal{Y}_{W}^{com})_{\dot{W}})$.

Algorithm ends.

Note that the algorithm terminates, the first time the affine space $w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u$ is contained in $(\text{sp}\{w_W^i, i = 0, 1, 2, \dots, j\} + \mathcal{Y}_{W}^{com})_{\dot{W}}$. This is when $j = k - 1$. Thus we have that, $(\mathcal{Y}_{W})_{\dot{W}} \equiv (\text{sp}\{w_W^j, j = 0, 1, 2, \dots, k-1\} + \mathcal{Y}_{W}^{com})_{\dot{W}}$ contains $w_{\dot{W}}^{k-1} + \mathcal{Y}_{W\dot{W}}^u \times \dot{W}$ and therefore also $\mathcal{Y}_{W\dot{W}}^u \times \dot{W}$. For $j < k - 1$, this does not happen.

Every vector in $w_{\dot{W}}^{j+1} + \mathcal{Y}_{W\dot{W}}^u \times \dot{W} = w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u, j = 0, 1, 2, \dots, k-1$, belongs to $(\mathcal{Y}_{W})_{\dot{W}}$ and

$$ \mathcal{Y}_{W}^{com} \leftrightarrow \mathcal{Y}_{W\dot{W}}^{u} \subseteq \mathcal{Y}_{W\dot{W}}^{u} \times \dot{W} \subseteq (\mathcal{Y}_{W})_{\dot{W}}. $$

Thus

$$ \mathcal{Y}_{W\dot{W}}^{u} \leftrightarrow \mathcal{Y}_{W} \subseteq (\mathcal{Y}_{W})_{\dot{W}}. $$

Now $\mathcal{Y}_{W\dot{W}}^u \circ W \supseteq \mathcal{Y}_{W}$, so that by Theorem 43,

$$ \mathcal{Y}_{W\dot{W}}^{u} \leftrightarrow (\mathcal{Y}_{W})_{\dot{W}} \supseteq \mathcal{Y}_{W}. $$

Thus $\mathcal{Y}_{W}$ is invariant in $\mathcal{Y}_{W\dot{W}}^u$.

Next, any invariant space in $\mathcal{Y}_{W\dot{W}}^u$ containing $(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$ must contain $w_W^0$, and therefore $(w_W^j \leftrightarrow \mathcal{Y}_{W\dot{W}}^u)_W, j = 0, \dots, k-1$. It follows therefore that it must contain $\mathcal{Y}_{W}$.

Thus, $\text{sp}\{w_W^j, j = 0, 1, 2, \dots, k-1\} + \mathcal{Y}_{W}^{com} = \mathcal{Y}_{W}$, the minimal invariant space of $\mathcal{Y}_{W\dot{W}}^u$ containing $(\mathcal{Y}_{W\dot{W}}^u \times \dot{W})_W$.

It is clear that $\mathcal{Y}_{W\dot{W}}^{start}$ is USG since $\mathcal{Y}_{W\dot{W}}^{start} \circ W = \mathcal{Y}_{W}$ and $\mathcal{Y}_{W\dot{W}}^{start} \circ \dot{W} \subseteq (\mathcal{Y}_{W})_{\dot{W}}$.

Next, the vectors $w_W^0, w_W^1, \dots, w_W^{k-1}$ are independent modulo $\mathcal{Y}_{W}^{com}$.