Transformation matrix of linear n-dimensional state-space system

[X89codered89X]

Hi all,

I have a linear algebra question relating actually to control systems (applied differential equations)

for the linear system

$<br /> {\dot{\vec{{x}}} = {\bf{A}}{\vec{{x}}} + {\bf{B}}}{\vec{{u}}}\\<br /> \\<br /> <br /> A \in \mathbb{R}^{ nxn }\\<br /> B \in \mathbb{R}^{ nx1 }\\$

In class, we formed a transformation matrix P using the controllability matrix $M_c$ as a basis (assuming it is full rank).
$<br /> M_c = [ {\bf{B \;AB \;A^2B\;...\;A^{n-1}B}}]$

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.

${\bf{M}}_2 = <br /> \begin{bmatrix}<br /> \alpha_{n-1} & \alpha_{n-2} & \cdots & \alpha_1 & 1 \\<br /> \alpha_{n-2} & \cdots & \alpha_1 & 1 & 0 \\<br /> \vdots & \alpha_1 & 1 & 0 & 0\\<br /> \alpha_1 & 1 & 0 & \cdots & 0\\<br /> 1 & 0 & 0& \cdots & 0 \\<br /> \end{bmatrix}<br />$

then the transformation matrix is then given by

$<br /> P^{-1} = M_c M_2<br />$and then applying the transformation always gives.. and this is what I don't understand...

${\overline{\bf{A}}} = {\bf{PAP}}^{-1} = <br /> <br /> \begin{bmatrix}<br /> 0 & 1 & 0 & \cdots & 0 \\<br /> 0 & 0 & 1 & \cdots & \vdots \\<br /> \vdots & \vdots & 0 & 1 & 0\\<br /> 0 & 0 & \cdots &0& 1\\<br /> -\alpha_{1} & -\alpha_{2} & \cdots & -\alpha_{n-1}& -\alpha_{n}\\<br /> \end{bmatrix}<br />$

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 ...

 

[X89codered89X]

Would I do better posting this somewhere else in PF?