# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 5 Classification of Lagrangian Systems (6)  
We start with a counterpart of Proposition 3. To formulate it, recall that the matrix \(E^{L}=M^{-1}P\), associated with the Lagrangian system (6), has real eigenvalues and is diagonalizable. It follows that, each eigenvalue \(\lambda_{j}\) of \(E_{d}^{L}\) is real and its eigenstructure is \(\Lambda_{j}=(\lambda_{j},\delta_{j},0)\), for \(1\leq j\leq q\), where \(q\) is the number of mutually distinct eigenvaluesof \(E_{d}^{L}\) and \(\mu_{j}=\delta_{j}\) are their algebraic multiplicities. In this case, we will simply denote \(\Lambda_{j}=(\lambda_{j},\mu_{j})\). Recall that, see Section 4, the matrix \(E_{d}\) represents the quotient map \(\hat{E}:\mathbb{R}^{n}/\mathcal{X}_{c}\to\mathbb{R}^{n}/\mathcal{X}_{c}\), induced by \(E:\mathbb{R}^{n}\to\mathbb{R}^{n}\), where \(\mathcal{X}_{c}=\mathcal{B}+\ldots+E^{k-1}\mathcal{B}\). **Proposition 4**. : _Two Lagrangian systems (6) and \((\widehat{\mathcal{L}\mathcal{S}})\), given by \((E^{L},B)\) and \((\hat{E}^{L},\hat{B})\), respectively, are LMF-equivalent if and only if their mechanical half-indices coincide, that is \(\hat{\mathcal{P}}(E^{L},B)=\hat{\mathcal{P}}(\hat{E}^{L},\hat{B})\) and, up to permutations, \((\lambda_{j},\mu_{j})=(\tilde{\lambda}_{j},\tilde{\mu}_{j})\), for \(1\leq j\leq q\), where \(\lambda_{j}\)'s and \(\tilde{\lambda}_{j}\)'s are distinct eigenvalues of \(E_{d}^{L}\) and \(\hat{E}_{d}^{L}\), respectively, and \(q\) is their number (the same for both systems)._  
A proof follows directly from Proposition 3 applied to two Lagrangian systems. The statements of Propositions 3 and 4 are formally the same; the only difference is that the eigenstructure of \(E_{d}\) of (2) can have any elements \(\Lambda_{j}\) (real or complex eigenvalues \(\lambda_{j}\) and diagonal or Jordan blocks), while \(\Lambda_{j}\) of \(E_{d}^{L}\) of (6) consists of real eigenvalues \(\lambda_{j}\) and diagonal blocks only. Contrary to Proposition 4, that is a Lagrangian counterpart of Proposition 3, the classification Theorem 2 does not apply to the Lagrangian systems (6) because, in general, the feedback transformation \(u=Fx+G\tilde{u}\) does not perserve the Lagrangian structure. Actually, the mechanical canonical form, given by \((E_{F},B_{F})\), is never Lagrangian because \(E_{F}\) has all eigenvalues \(\lambda_{j}=0\), and thus never can be written as a product of two symmetric matrices \(E_{F}=-M^{-1}P\), for \(M\) being invertible. Indeed, a symmetric matrix, whose all eigenvalues are zero, is \(P=0\) but then \(-M^{-1}P=0\
eq E_{F}\). Instead of Theorem 2, we have the following result. **Proposition 5**. : _Any Lagrangian mechanical system (6) is LMF-equivalent to the following Lagrangian system:_  
\[ \begin{array}{ll}\dot{x}_{c}=y_{c}&\dot{x}_{d}=y_{d}\\ \dot{y}_{c}=E_{F}^{L}x_{c}+B_{F}u&\dot{y}_{d}=E_{d}^{L}x_{d},\end{array}  \tag{19}\]  
_where \(E_{d}^{L}=\operatorname{diag}(\lambda_{1}^{d},\ldots,\lambda_{l}^{d})\), \(l=n-k\), with \(\lambda_{j}^{d}\) arbitrary real, and \(E_{F}^{L}\) is of the same form as \(E_{F}\), given by (16) and (17), with \(N_{i}\) replaced by \(L_{i}= \begin{pmatrix}0&1&\ldots&0\\ &&\ddots&\\ 0&0&\ldots&1\\ a_{1}^{i}&a_{2}^{i}&\ldots&a_{p_{i}}^{i}\end{pmatrix} ,\) such that the eigenvalues of \(L_{i}\) are all real and mutually distinct._
2. _Any two Lagrangian systems of the form (_19_), with the same_ \(E_{d}^{L}\) _but arbitrary terms_ \(a_{j}^{i}\) _and_ \(\dot{a}_{j}^{i}\) _(such that, for each block_ \(1\leq i\leq m\)_, the eigenvalues_ \(\lambda_{j}^{i}\) _of_ \(L_{i}\)_are mutually distinct and so are_ \(\tilde{\lambda}_{j}^{i}\) _of_ \(L_{i}\)_), are LMF-equivalent._  
Proof. : \((i)\). By Theorem 2, any (6) is LMF-equivalent to (18), whose \(x_{c}\)-subsystem is not Lagrangian, for which \(E_{d}^{l}\) is \(E_{d}^{L}\) (since all eigenvalues of (6) are real).