Primary Decomposition Theorem

== Theorem ==
Let $K$ be a field.
Let $V$ be a vector space over $K$.
Let $T: V \to V$ be a linear operator on $V$.
Let $\map p x \in K \sqbrk x$ be a polynomial such that:
:$\map \deg p \ge 1$
:$\map p T = 0$
where $0$ is the zero operator on $V$.
Let $\map {p_1} x, \map {p_2} x, \ldots, \map {p_r} x$ be distinct irreducible monic polynomials.
Let $c \in K \setminus \set 0$ and $a_1, a_2, \ldots, a_r, r \in \Z_{\ge 1}$ be constants.
We have that:
:$\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \dotsm \map {p_r} x^{a_r}$
The primary decomposition theorem then states the following :
:$(1): \quad \map \ker {\map {p_i} T^{a_i} }$ is a $T$-invariant subspace of $V$ for all $i = 1, 2, \dotsc, r$
:$(2): \quad \ds V = \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$
== Proof of $(1)$ ==
Let $v \in \map \ker {\map {p_i} T^{a_i} }$.
Then:
This shows that:
:$\map T v \in \map \ker {\map {p_i} T^{a_i} }$
for all $v \in \map \ker {\map {p_i} T^{a_i} }$.
This is because $v$ was first arbitrary in $\map \ker {\map {p_i} T^{a_i} }$.
That is:
:$\map \ker {\map {p_i} T^{a_i} }$
is a $T$-invariant subspace of $V$.
== Proof of $(2)$ ==
Proof by induction on $r$:
For all $r \in \Z_{\ge 1}$, let $\map P r$ be the proposition:
:$V = \ds \bigoplus_{i \mathop = 1}^r \map \ker {\map {p_i} T^{a_i} }$
where $\map {p_i} x^{a_i}$ for $i = 1, 2, \dotsc, r$ satisfy the hypotheses of the theorem.
$V$ and $T$ are arbitrary.
=== Basis for the Induction ===
We have:
Thus $\map P 1$ holds.
This is our basis for the induction
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$V = \ds \bigoplus_{i \mathop = 1}^{k - 1} \map \ker {\map {p_i} T^{a_i} }$
where $\map {p_i} x^{a_i}$ for $i = 1, 2, \dotsc, r$ satisfy the hypotheses of the theorem.
$V$ and $T$ are arbitrary.
Then we need to show:
:$V = \ds \bigoplus_{i \mathop = 1}^k \map \ker {\map {p_i} T^{a_i} }$
=== Induction Step ===
Suppose we meet the hypotheses of the theorem when $r = k$.
Then:
:$\map p x \in K \sqbrk x$
is a polynomial such that:
:$\map \deg p \ge 1$
:$\map p T = 0$
:$\map p x = c \map {p_1} x^{a_1} \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$
Let $W := \map \ker {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} }$.
First it will be shown that $W$ is a $T$-invariant subspace of $V$.
Let $w \in W$.
Then:
This shows that $\map T w \in W$ for all $w \in W$, because $w$ was first arbitrary in $W$.
That is, $W$ is a $T$-invariant subspace of $V$.
So it makes sense to talk about the restriction $T \restriction_W$ of $T$ on $W$.
Now, define:
:$\map q x := \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$
Let $w \in W$.
Then:
Because $w$ was arbitrary in $W$, this shows that $\map q {T \restriction_W} = 0$, where $0$ is the zero operator on $W$.
The hypotheses of the theorem are met.
So, by the induction hypothesis:
:$\ds W = \bigoplus_{i \mathop = 2}^k \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$
Now we show that:
:$\map \ker {\map {p_i} {T \restriction_W}^{a_i} } = \map \ker {\map {p_i} T^{a_i} }$
for all $i = 2, \dotsc, k$.
Thus, for all $i = 2, \dotsc, k$:
Then:
:$\map \ker {\map {p_i} T^{a_i} } \subseteq W$
for all $i = 2, \dotsc, k$.
Let:
:$v \in \map \ker {\map {p_j} T^{a_j} }$
where $j \in \set {2, \dotsc, k}$.
Then:
So:
:$v \in \map \ker {\map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } = W$
and:
:$\map \ker {\map {p_j} T^{a_j} } \subseteq W$
because $v$ was arbitrary in $\map \ker {\map {p_j} T^{a_j} }$.
So:
:$v \in \map \ker {\map {p_i} T^{a_i} } \implies v \in W$
and so:
:$\map {\map {p_i} {T \restriction_W}^{a_i} } v = \map {\map {p_i} T^{a_i} } v = 0$
for all $i = 2, \dotsc, k$.
Finally, because $v$ was arbitrary in $\map \ker {\map {p_i} T^{a_i} }$:
:$v \in \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$
and:
:$\map \ker {\map {p_i} T^{a_i} } \subseteq \map \ker {\map {p_i} {T \restriction_W}^{a_i} }$
for all $i = 2, \dotsc, k$.
This shows that:
:$\map \ker {\map {p_i} {T \restriction_W}^{a_i} } = \map \ker {\map {p_i} T^{a_i} }$
for all $i = 2, \dotsc, k$.
So we conclude that:
:$\ds W = \bigoplus_{i \mathop = 2}^k \map \ker {\map {p_i} T^{a_i} }$
In order to show that:
:$V = \ds \bigoplus_{i \mathop = 1}^k \map \ker {\map {p_i} T^{a_i} }$
we equivalently show that:
:$(a): \quad V = W + \map \ker {\map {p_1} T^{a_1} }$
:$(b): \quad W \cap \map \ker {\map {p_1} T^{a_1} } = 0$
where $0 = \set 0 \subseteq V$.
Notice that $(b)$ is equivalent to showing that:
:$\map \ker {\map {p_1} T^{a_1} } \cap \map \ker {\map {p_i} T^{a_i} } = 0$
for all $i = 2, \dotsc, k$.
In particular, $(b)$ implies this last result, because $\map \ker {\map {p_i} T^{a_i} } \subseteq W$ for all $i = 2, \dotsc, k$.
$(a): \quad$ We have that:
:$\map g x := \gcd \set {\map {p_1} x^{a_1}, \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} } = 1 \in K$.
Indeed, $\map {p_1} x$ being an irreducible monic polynomial, either $\map g x = 1$ or $\map g x = \map {p_1} x^z$ for some $1 \le z \le a_1$.
But we know that the greatest common divisor of two polynomials divides both of these polynomials.
So we must conclude that $\map g x = 1$ because otherwise we would have $\map {p_1} x = \map {p_i} x$ for some $i \in \set {2, \ldots, k}$.
This is a contradiction with the hypothesis that the $\map {p_i} x$ ($i = 1, 2, \ldots, k$) are all distinct.
So, by Bézout's_Identity for polynomials, we know that:
:$\exists \map h x, \map {h_1} x \in K \sqbrk x$
such that:
:$1 = \map {h_1} x \map {p_1} x^{a_1} + \map h x \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k}$
Let $v \in V$.
Then, evaluating the last polynomial equality at $T$, and evaluating the resulting operator at $v$, we obtain:
:$v = \map {\map {h_1} T \map {p_1} T^{a_1} } v + \map {\map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } v$
Notice that $\map I v = v$ on the .
But:
So:
:$\map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} \in \map \ker {\map {p_1} T^{a_1} }$
Similarly, we find that:
:$\map {\map {p_2} T^{a_2} \map {p_3} T^{a_3} \cdots \map {p_k} T^{a_k} } {\map {\map {h_1} T \map {p_1} T^{a_1} } v} = 0$
and so:
:$\map {\map {h_1} T \map {p_1} T^{a_1} } v \in \map \ker {\map {p_2} T^{a_2} \map {p_3} T^{a_3} \cdots \map {p_k} T^{a_k} } = W$
Because $v$ was arbitrary in $V$:
:$v = \map {h_1} T \map {p_1} T^{a_1} v + \map h T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} v \implies V = W + \map \ker {\map {p_1} T^{a_1} }$
$(b): \quad$ Define:
:$S := W \cap \map \ker {\map {p_1} T^{a_1} }$
:$M := \set {\map f x \in K \sqbrk x: \map {\map f T} s = 0 \forall s \in S}$
and:
:$\map q x := \map \gcd M$
By definition of $W$:
:$\map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} \in M$
As every element of $S$ is also an element of $\map \ker {\map {p_1} T^{a_1} }$:
:$\map {p_1} x^{a_1} \in M$
it follows that:
:$\map q x \divides \map {p_1} x^{a_1}$
and:
:$\map q x \divides \map {p_2} x^{a_2} \dotsm \map {p_k} x^{a_k}$
We know that the greatest common divisor $\map {u_1} x$ of two polynomials $\map {u_2} x$ and $\map {u_3} x$ has the property that:
:if a polynomial $\map {u_4} x$ divides both $\map {u_2} x$ and $\map {u_3} x$
:then $\map {u_4} x$ must also divide $\map {u_1} x$.
Thus, we must have:
:$\map q x \divides \gcd \set {\map {p_1} x^{a_1}, \map {p_2} x^{a_2} \cdots \map {p_k} x^{a_k} } = \map g x = 1$
So:
:$\map q x = \map g x = 1$
Bézout's_Identity for polynomials may then be applied to obtain:
:$\map q x = 1 = \map {h_1} x \map {p_1} x^{a_1} + \map {h_2} x \map {p_2} x^{a_2} \dotsm \map {p_k} x^{a_k}$
for some $\map {h_1} x, \map {h_2} x \in K \sqbrk x$.
Let $s \in S$.
Then, evaluating the last polynomial equality at $T$, and evaluating the resulting operator at $s$, we obtain, by definition of $M$ and by noticing that the then becomes $\map I s = s$:
:$s = \map {\map {h_1} T \map {p_1} T^{a_1} } s + \map {\map {h_2} T \map {p_2} T^{a_2} \cdots \map {p_k} T^{a_k} } s = 0$
Because $s$ was arbitrary in $S$, it follows that:
:$S = W \cap \map \ker {\map {p_1} T^{a_1} } = 0$
 
== Remarks ==
$V$ need not be of finite dimension, but if it is then we can see $T$ as its matrix in the equations and the notation $T^j$ is then directly related to matrix multiplication.
By definition, when applying $\map p x$ at $T$, one needs to convert constants which may appear in the expression of $\map p x$ into transformations in the following way
For example, if $\map p x = 2 + x$, then $\map p T = 2 I + T$, where $I$ is the identity application $I: V \to V$.
$\map {p_1} x, \map {p_2} x, \dotsc, \map {p_r} x$ are not necessarily of degree $1$.
For example, if $K = \R$ and $\map p x = x^2 + 1$, then $\map p x$ is irreducible on $\R$ and $\map {p_1} x = \map p x$ is of degree $2$.
$T^j \triangleq \begin{cases}
\overbrace {T \circ T \circ \cdots \circ T}^{\text {$j$ times} } & : j \in \Z_{\ge 1} \\
I: V \to V & : j = 0
\end{cases}$
By definition, the greatest common divisor of two polynomials is a monic polynomial.
Category:Polynomial Theory
Category:Named Theorems