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Hi all,

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^{n-1}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^{n-1} +... + \alpha_{n-1}\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_{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}

[/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_{n-1}& -\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...

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

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^{n-1}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^{n-1} +... + \alpha_{n-1}\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_{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}

[/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_{n-1}& -\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...

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

Last edited: