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Linear System Transformations 
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#1
Nov2112, 08:02 PM

P: 137

Hi there,
I have a linear algebra question relating actually to control systems (applied differential equations) for the linear system [itex] {\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\ \\ A \in \mathbb{R}^{ nxn }\\ B \in \mathbb{R}^{ nx1 }\\ [/itex] In class, we formed a transformation matrix P using the controllability matrix [itex] M_c [/itex] as a basis (assuming it is full rank). [itex] M_c = [ {\bf{B \;AB \;A^2B\;....\;A^{n1}B}}] [/itex] and there is a second matrix with a less established name. Given that the characteristic equation of the system is [itex] I\lambda A = \lambda^n + \alpha_1 \lambda^{n1} +... + \alpha_{n1}\lambda + \alpha_n= 0 [/itex], we then construct a second matrix, call it M_2, which is given below. [itex] {\bf{M}}_2 = \begin{bmatrix} \alpha_{n1} & \alpha_{n2} & \cdots & \alpha_1 & 1 \\ \alpha_{n2} & \cdots & \alpha_1 & 1 & 0 \\ \vdots & \alpha_1 & 1 & 0 & 0\\ \alpha_1 & 1 & 0 & \cdots & 0\\ 1 & 0 & 0& \cdots & 0 \\ \end{bmatrix} [/itex] then the transformation matrix is then given by [itex] P^{1} = M_c M_2 [/itex] and then applying the transformation always gives.. and this is what I don't understand.... [itex] {\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_{n1}& \alpha_{n}\\ \end{bmatrix} [/itex] 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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