# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 6 Stability and Stabilization  
Now we come back to (not necessarily Lagrangian) mechanical systems (2) and we turn our attention to the relation between the eigenvalues of \(E\), denoted \(\lambda_{j}\), and the eigenvalues of \(\hat{A}\), denoted \(\sigma_{j}\), since the latter are responsible for the stability of the system. Let \(\hat{A}=\left( \begin{array}{cc}0&I_{n}\\ E&0\end{array} \right)\). If \(\lambda\) is an eigenvalue of \(E\), then \(\sigma=\pm\sqrt{\lambda}\) are eigenvalues of \(\hat{A}\). By a direct calculation, we see that the characteristic polynomial of \(\hat{A}\) is  
\[\det(\sigma I_{2n}-\hat{A})= \det\biggl{(} \begin{pmatrix}\sigma I_{n}&0\\ 0&\sigma I_{n}\end{pmatrix} - \begin{pmatrix}0&I_{n}\\ E&0\end{pmatrix} \biggr{)}=\det(\sigma^{2}I_{n}-E)\] \[=(\sigma^{2}-\lambda_{1})(\sigma^{2}-\lambda_{2})\ldots(\sigma^ {2}-\lambda_{n}),\]  
since all blocks of \(\hat{A}\) commute, and where \(\lambda_{1}\ldots\lambda_{n}\) are the eigenvalues of \(E\). The above polynomial has the roots \(\sigma_{j}=\pm\sqrt{\lambda_{j}}\). If \(\lambda\in\mathbb{R}\), then \(\sigma=\pm a\) or \(\sigma=\pm a\texttt{i}\), where \(a=\sqrt{|\lambda|}\in\mathbb{R}_{+}\). If \(\lambda=\alpha+\beta\texttt{i}\in\mathbb{C}\) (and its conjugate \(\alpha-\beta\texttt{i}\in\mathbb{C}\)), then \(\sigma=\pm(a+b\texttt{i})\) and \(\pm(a-b\texttt{i})\), where \(\alpha=a^{2}-b^{2}\) and \(\beta=2 ab\). Denote \(\mathbb{R}_{\leq 0}=\{\lambda\in\mathbb{R}:\lambda\leq 0\}\). Let us visualize the relation between the eigenvalues with the following table. \begin{tabular}{c c c} \hline \hline \(\lambda(E)\) & \(\sigma(\hat{A})\) & **Sketch** \\ \hline \(\mathbb{R}_{+}\) & \(\pm a\) & \\ \hline \(\mathbb{R}_{\leq 0}\) & \(\pm a\texttt{i}\) & \\ \hline \(\mathbb{C}\) & \(\pm(a+b\texttt{i})\) and \(\pm(a-b\texttt{i})\) & \\ \hline \end{tabular}   
The above analysis of the eigenvalues leads to the following simple but important observation. We will say that a dynamical system \(\dot{z}=\hat{A}z\), for \(\hat{A}\) given by (4) and \(z=(x,y)\), is stable if all its equilibrium points are stable. Recall that \(\mu_{j}\) denotes the algebraic multiplicity of an eigenvalue \(\lambda_{j}\) and \(\gamma_{j}=\dim\ker(\lambda_{j}I_{n}-E)\) its geometric multiplicity. Proposition 7: _Consider the dynamical system \(\dot{z}=\hat{A}z\), where \(\hat{A}= \begin{pmatrix}0&I_{n}\\ E&0\end{pmatrix} \) or, equivalently, \(\ddot{x}=Ex\)._  
1. _The system_ \(\dot{z}=\hat{A}z\) _is never asymptotically stable._
2. _The following conditions are equivalent:_ 1. _the system_ \(\dot{z}=\hat{A}z\) _is stable,_ 2. _all eigenvalues_ \(\lambda_{j}\) _of_ \(E\) _satisfy_ \(\lambda_{j}\in\mathbb{R}_{\leq 0}\) _and, moreover, their algebraic and geometric multiplicities coincide, i.e.,_ \(\mu_{j}=\gamma_{j}\)_, for_ \(1\leq j\leq q\)_, where_ \(q\) _is the number of distinct eigenvalues of_ \(E\)_,_ 3. _E has a Lagrangian structure, i.e.,_ \(E=M^{-1}P\) _for some symmetric matrices_ \(M\) _(invertible) and_ \(P\) _and the eigenvalues of_ \(E\) _satisfy_ \(\lambda_{j}\in\mathbb{R}_{\leq 0}\)_._  
Proof: It is immediate to see \((AS)\), since at least for one eigenvalue \(\sigma_{j}\) of \(\hat{A}\) we have \(\operatorname{Re}\sigma_{j}\geq 0\). Equivalence of \((i)\iff(ii)\) follows immediately from Lemma 2 and the table above. For a proof of necessity of \(\mu_{j}=\gamma_{j}\) (which is equivalent to rank \((\lambda_{j}I_{n}-E)=n-\mu_{j}\)), see the proof of Theorem 4.5 in [16]. Equivalence \((ii)\iff(iii)\) is given by Proposition 1 and the comments below it. Of course, the class of second-order differential equations (or dynamical systems) \(\ddot{x}=Ex\) is bigger than the class of Lagrangian systems \(\ddot{x}=E^{L}x\) because the matrix \(E\) can be any, while \(E^{L}\) has to be \(\mathbb{R}\)-diagonalizable. It is obvious that \(\ddot{x}=E^{L}x\) cannot be asymptotically stable because it preserves the energy \(\frac{1}{2}\dot{x}^{T}M\dot{x}+\frac{1}{2}\dot{x}^{T}Px\).