Linear System Transformations

by X89codered89X
Tags: control systems, linear algebra, odes
 P: 136 Hi there, I have a linear algebra question relating actually to control systems (applied differential equations) for the linear system ${\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\ \\ A \in \mathbb{R}^{ nxn }\\ B \in \mathbb{R}^{ nx1 }\\$ In class, we formed a transformation matrix P using the controllability matrix $M_c$ as a basis (assuming it is full rank). $M_c = [ {\bf{B \;AB \;A^2B\;....\;A^{n-1}B}}]$ and there is a second matrix with a less established name. Given that the characteristic equation of the system is $|I\lambda -A| = \lambda^n + \alpha_1 \lambda^{n-1} +... + \alpha_{n-1}\lambda + \alpha_n= 0$, we then construct a second matrix, call it M_2, which is given below. ${\bf{M}}_2 = \begin{bmatrix} \alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\ \alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\ \vdots & \alpha_1 & 1 & 0 & 0\\ \alpha_1 & 1 & 0 & \cdots & 0\\ 1 & 0 & 0& \cdots & 0 \\ \end{bmatrix}$ then the transformation matrix is then given by $P^{-1} = M_c M_2$ and then applying the transformation always gives.. and this is what I don't understand.... ${\overline{\bf{A}}} = {\bf{PAP}}^{-1} = \begin{bmatrix} 0 & 1 & 0 & \cdots & 0 \\ 0 & 0 & 1 & \cdots & \vdots \\ \vdots & \vdots & 0 & 1 & 0\\ 0 & 0 & \cdots &0& 1\\ -\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\ \end{bmatrix}$ Now I'm just looking for intuition is to why this is true. I know that this only works if the controllability matrix is full rank, which can the be used as a basis for the new transformation, but I don't get how exactly the M_2 matrix is using it to transform into the canonical form.... Can someone explain this to me? thanks... Disclaimer: I posted this in another PF subforum, but I think I might do better in this section.

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