# A Mechanical Feedback Classification of Linear Mechanical Control Systems
## 2 Problem Statement  
### Linear Mechanical Control Systems  
In this subsection, we introduce the object of our study, namely linear mechanical control systems. This class is larger than linear Lagrangian control systems that will also be introduced in this subsection and form an important subclass in our study. Consider the linear mechanical control system with \(n\) degrees of freedom and \(m\) controls  
\[\ddot{x}=Ex+Bu, \tag{1}\]  
where \(x\in\mathbb{R}^{n}\) are the configurations (the generalized coordinates). The matrix \(E\) is an \(n\times n\) constant real matrix corresponding to an uncontrolled (depending on configurations only but, possibly, non potential) force in the system and the input matrix \(B\) is an \(n\times m\) constant real matrix describing external forces controlled by the controls \(u\in\mathbb{R}^{m}\). Equivalently, system (1) can be represented as a first-order system on the state-space \(\mathbb{R}^{n}\times\mathbb{R}^{n}\), equipped with coordinates \((x,y)\) denoting, respectively, configurations and velocities  
\[ \begin{array}{ll}\mathcal{LMS}&\dot{x}=y\\ &\dot{y}=Ex+Bu,\end{array}  \tag{2}\]  
or as a linear control system of dimension \(2n\), with coordinates \(z=(x,y)\in\mathbb{R}^{n}\times\mathbb{R}^{n}\) (to be precise \(z\) is a \"stacked\" vector \(z=(x^{T},y^{T})^{T}\), however we decide to skip this notation for a clarity sake), given by, compare [9]  
\[\dot{z}=\hat{A}z+\hat{B}u, \tag{3}\]where  
\[\hat{A}=\left( \begin{array}{cc}0&I_{n}\\ E&0\end{array} \right),\qquad\hat{B}=\left( \begin{array}{c}0\\ B\end{array} \right). \tag{4}\]  
Notice that the equation \(\dot{y}=Ex+Bu\) is not of the more general form \(\dot{y}=Ex+Ly+Bu\), that is, contains neither gyroscopic nor dissipative terms \(Ly\), see e.g., [6; 7; 9; 10] for a discussion of both classes. Our obvious inspiration are linear Lagrangian (conservative) systems which constitute a subclass of (2) whose configuration space is the real vector space \(\mathbb{R}^{n}\), equipped with an inner product on the space \(\mathbb{R}^{n}\) of velocities given by a real valued quadratic form \(\frac{1}{2}\hat{x}^{T}M\hat{x}\) (describing the kinetic energy of the system), where \(M\) is a constant real symmetric and positive definite matrix (\(M^{T}=M>0\)). Moreover, we consider the potential energy given by a quadratic form \(V=\frac{1}{2}x^{T}Px\), where \(P\) is a symmetric potential matrix (\(P^{T}=P\)). The corresponding quadratic Lagrangian reads \(\mathcal{L}=\frac{1}{2}\hat{x}^{T}M\hat{x}-\frac{1}{2}x^{T}Px\). The derivation of the Euler-Lagrange equations yields the second-order system:  
\[M\hat{x}+Px=Ku, \tag{5}\]  
where \(K\) is an \(n\times m\) real matrix whose columns are vectors corresponding to the external controlled forces. Straightforward calculations show that any Lagrangian linear mechanical control system can be represented by a particular form of (2), namely  
\[ \begin{array}{ll}\mathcal{LLS}&\dot{x}=y\\ &\dot{y}=E^{L}x+Bu,\end{array} \quad\text{where }E^{L}=-M^{-1}P\text{ and }B=M^{-1}K. \tag{6}\]  
Notice that for Lagrangian systems (6) the matrix \(B\) can be any (since \(K\) is arbitrary) but the matrix \(E^{L}\) is special, namely the product of two symmetric matrices (the first being invertible), which we call the Lagrangian structure. Another characterization of (6) can be formulated as follows [11]. **Proposition 1**. : _The linear mechanical control system (2) is a Lagrangian system of the form (6) if and only if there exists a real invertible map \(T:\mathbb{R}^{n}\to\mathbb{R}^{n}\) that diagonalizes the matrix \(E\), that is, \(TET^{-1}=\operatorname{diag}(\lambda_{1},\dots,\lambda_{n})\)._  
For a proof see [11]. The above proposition can be rephrased as follows. All distinct eigenvalues \(\lambda_{j}\) of the matrix \(E\), \(1\leq j\leq q\), where \(q\) is the number of distinct eigenvalues, are real, i.e., \(\lambda_{j}\in\mathbb{R}\), and there are no Jordan blocks.