# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 3 Classification of Controllable Systems (2)  
A linear mechanical control system (2) is _controllable_ if for any \(t_{0}\), any initial state \(z_{0}=(x_{0},y_{0})\), and any final state \(z_{f}=(x_{f},y_{f})\) there exist \(t_{f}>t_{0}\) and a control \(u:[t_{0},t_{f}]\rightarrow\mathbb{R}^{m}\), such that \(z(t_{0})=(x_{0},y_{0})\) and \(z(t_{f})=(x_{f},y_{f})\). The following result is a straightforward generalization of [10], see also [9] and cf. a classical work on modal controllability for Lagrangian class including dissipative forces [7; 14]. **Lemma 1** (Controllability of (2)). : _For (2) the following statements are equivalent:_  
1. \((\mathcal{L}\mathcal{M}\mathcal{S})\) _is controllable_
2. \(\mathrm{rank}\left(\hat{B},\hat{A}\hat{B},\ldots,\hat{A}^{2n-1}\hat{B}\right)= 2n\) _(Kalman Rank Condition)__(iii)_ \(\text{rank }\left(\hat{B},\hat{A}^{2}\hat{B},\ldots,\hat{A}^{2(n-1)}\hat{B} \right)=n\)__  
_(iv)_ \(\text{rank }\left(B,EB,\ldots,E^{n-1}B\right)=n\) _(Mechanical Kalman Rank Condition)_  
Proof. : From the Kalman controllability result, we have \((i)\iff(ii)\). The rest of the proof follows from a direct computation of the Kalman controllability matrix  
\[\left(\hat{B},\hat{A}\hat{B},\ldots,\hat{A}^{2n-1}\hat{B}\right)=\left( \begin{array}{cccccc}0&B&0&EB&\ldots&0&E^{n-1}B\\ B&0&EB&0&\ldots&E^{n-1}B&0\end{array} \right). \tag{10}\]  
Therefore we see that we can take only even powers \(\hat{A}^{2i}\hat{B}\) in \((iii)\) or the lower part of the matrix (10) as in \((iv)\). **Remark 2**. : _Note that the Machanical Kalman Rank Condition, i.e., item \((iv)\) of Lemma 1, uses objects on the configuration space \(\mathbb{R}^{n}\) only, while the state-space of (2) is \(\mathbb{R}^{n}\times\mathbb{R}^{n}\). Apart of a pure mathematical value, that sort of reduction is practically motivated since computations are simpler. All of our further results share that property._  
Now attach to the system (2) an \(n\)-tuple of indices \((\bar{r}_{0},\ldots,\bar{r}_{n-1})\)  
\[ \begin{split}\bar{r}_{0}&=\text{rank}(B),\\ \bar{r}_{i}&=\text{rank}\Big{(}B,EB,...,E^{i}B\Big{)}- \text{rank}\Big{(}B,EB,...,E^{i-1}B\Big{)},\end{split}  \tag{11}\]  
for \(1\leq i\leq n-1\). Furthermore, define the dual indices  
\[\bar{\rho}_{j}=\text{card}(\bar{r}_{i}\geq j:0\leq i\leq n-1)\qquad\text{for }1 \leq j\leq m. \tag{12}\]  
These integers, which we will call _mechanical half-indices_, are mechanical analogues of the controllability (Brunovsky, Kronecker) indices \(r_{i}\)'s and \(\rho_{j}\)'s of system (3) defined by (11) and (12), respectively, with \(E\) replaced by \(\hat{A}\) and \(B\) by \(\hat{B}\). Note that the indices \(\bar{\rho}_{i}\) are invariant under (7) and (8), therefore they form a set of invariants attached to (2). Actually, they form a set of complete invariants, as we will show in Theorem 1 below. We denote the above sequences as \(\mathcal{R}(E,B)=(\bar{r}_{0},\ldots,\bar{r}_{n-1})\) and \(\bar{\mathcal{P}}(E,B)=(\bar{\rho}_{1},\ldots,\bar{\rho}_{m})\). **Proposition 2**. : _For the mechanical control system \((\mathcal{LMS})\)_  
1. _the sequence of indices_ \(\mathcal{R}\big{(}\hat{A},\hat{B}\big{)}=(r_{0},r_{1},\ldots,r_{2n-1})\) _is the doubled sequence of_ \(\mathcal{R}(E,B)\)_, i.e.,_ \((r_{0},r_{1},\ldots,r_{2n-1})=(\bar{r}_{0},\bar{r}_{0},\bar{r}_{1},\bar{r}_{ 1},\ldots,\bar{r}_{n-1},\bar{r}_{n-1})\)_;_
2. _the mechanical half-indices are half of the controllability indices, i.e.,_ \(\rho_{j}=2\bar{\rho}_{j}\)_, for_ \(1\leq j\leq m\)_._  
Proof. : Let us invoke the controllability matrix of \((\mathcal{LMS})\) given by (10) and calculate \(r_{0}=\text{rank }\hat{B}\) and \(r_{i}=\text{rank }\big{(}\hat{B},\ldots,\hat{A}^{i}\hat{B}\big{)}-\text{rank } \big{(}\hat{B},\ldots,\hat{A}^{i-1}\hat{B}\big{)}\), and compare them with \(\bar{r}_{i}\) given by (11). The crucial observation is that we can calculate the ranks of the lower and upper submatrices separately and then add them. \[