Transformation matrix of linear n-dimensional state-space system

1. Nov 15, 2012

X89codered89X

Hi all,

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...

All Right! I think I'm done editin LATEX ....

Last edited: Nov 15, 2012
2. Nov 16, 2012

X89codered89X

Would I do better posting this somewhere else in PF?