# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 4 Classification of Uncontrollable Systems (2)
We have \(E\mathcal{X}_{c}\subset\mathcal{X}_{c}\) (since \(\mathrm{rank}\;(B,EB,\ldots,E^{k-1}B)=k\)), so \(E\) gives rise to a well defined map \(\tilde{E}:\mathbb{R}^{n}/\mathcal{X}_{c}\to\mathbb{R}^{n}/\mathcal{X}_{c}\) between the quotient spaces (i.e., factor spaces). Choose a subspace \(\mathcal{X}_{d}\subset\mathbb{R}^{n}\) such that \(\mathcal{X}_{c}\oplus\mathcal{X}_{d}=\mathbb{R}^{n}\) and any linear coordinates \((x_{c},x_{d})\) on \(\mathbb{R}^{n}\) such that \(\mathcal{X}_{c}=\left\{ \begin{pmatrix}x_{c}\\ 0\end{pmatrix} :x_{c}\in\mathbb{R}^{k}\right\}\) and \(\mathcal{X}_{d}=\left\{ \begin{pmatrix}0\\ x_{d}\end{pmatrix} :x_{d}\in\mathbb{R}^{n-k}\right\}\). Thus the matrix \(E\) takes in the \((x_{c},x_{d})\)-coordinates the block-triangular form \(E= \begin{pmatrix}E_{c}^{1}&E_{c}^{2}\\ 0&E_{d}\end{pmatrix} \) and the map \(\tilde{E}\) in these coordinates is given by the matrix \(E_{d}\). Choosing another subspace \(\tilde{\mathcal{X}}_{d}\) (completing \(\mathcal{X}_{c}\) to \(\mathbb{R}^{n}\)) will lead to another system of coordinates \((\tilde{x}_{c},\tilde{x}_{d})\) and a matrix \(\tilde{E}_{d}\) related to \(E_{d}\) by \(\tilde{E}_{d}=TE_{d}T^{-1}\). So the eigenvalues of \(\tilde{E}\) are well defined as those of \(E_{d}\) (being, obviously, the same as those of \(\tilde{E}_{d}\)). Let \(\lambda_{1},\ldots,\lambda_{q}\) be \(q\) mutually distinct eigenvalues of the matrix \(E_{d}\). For each \(\lambda_{j}\), \(1\leq j\leq q\), let \(\delta_{j}\) be the dimension of the diagonal block of \(\lambda_{j}\) and let \(\kappa_{j}\) be the number of Jordan blocks of \(\lambda_{j}\), whose dimensions are \(\epsilon_{j}^{1},\ldots,\epsilon_{j}^{\kappa_{j}}\), respectively. We define the eigenstructure \(\Lambda_{j}\) of \(\lambda_{j}\) by \(\Lambda_{j}=(\lambda_{j},\delta_{j},\epsilon_{j}^{1},\ldots,\epsilon_{j}^{ \kappa_{j}})\). Clearly, the same eigenvalue \(\lambda_{j}\) may appear in more than one block and the algebraic multiplicity of \(\lambda_{j}\) is \(\mu_{j}=\delta_{j}+\epsilon_{j}^{1}+\ldots+\epsilon_{j}^{\kappa_{j}}\). Proposition 3: _Two mechanical control systems (2) and \((\widetilde{\mathcal{LMS}})\) given by \((E,B)\) and \((\tilde{E},\tilde{B})\), respectively, are LMF-equivalent if and only if their mechanical half-indices coincide, that is \(\tilde{\mathcal{P}}(E,B)=\tilde{\mathcal{P}}(\tilde{E},\tilde{B})\), and the eigenstructures of \(E_{d}\) and \(\tilde{E}_{d}\) coincide \((\Lambda_{1},\ldots,\Lambda_{q})=(\tilde{\Lambda}_{1},\ldots,\tilde{\Lambda}_ {q})\), up to a permutation of \(\Lambda_{j}\)'s, where \(q\), the number of distinct eigenvalues, is the same for both systems._  
Proof: By Theorem 2, both systems can be brought, by LMF-transformations, into their canonical form (18). Then, both controllable subsystems are LMF-equivalent if and only if their mechanical half-indices coincide (see Theorem 1). Finally, if the eigenstructures of \(E_{d}^{I}\) and \(\tilde{E}_{d}^{I}\) coincide, then they are the same matrices up to a permutation of blocks (note that both are in the Jordan form) and vice versa. Thus the composition of both LMF-transformations converts (2) into \((\widetilde{\mathcal{LMS}})\). \(\qed\)  
Example 1: _Consider two systems (2) and \((\widetilde{\mathcal{LMS}})\), with \(n=3\), \(m=1\) represented, respectively, by the following pairs \((E,B)\) and \((\tilde{E},\tilde{B})\):_  
\[E= \begin{pmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{pmatrix} ,\quad\tilde{E}= \begin{pmatrix}1&1&0\\ 0&1&0\\ 0&0&1\end{pmatrix} ,\quad B=\tilde{B}= \begin{pmatrix}0\\ 0\\ 1\end{pmatrix} .\]  
_It can be easily checked that the two systems are not LMF-equivalent. For both systems \(k=1\) and the mechanical half-indices are \(\tilde{\rho}_{1}=1\).