I am given the two linear systems: \begin{eqnarray} \Sigma_1: \dot{x}&=&Ax+Bu\\ y&=&Cx+Du \end{eqnarray} and \begin{eqnarray} \Sigma_2: \dot{x}&=&\bar{A}x+\bar{B}u\\ y&=&\bar{C}x+\bar{D}u \end{eqnarray} with the same amount of inputs and outputs. Here $\bar{A} =SAS^{-1}$, $\bar{B} =SB$, $\bar{C}=CS^{-1}$ and $\bar{D}=D$ where $S$ is a non singular matrix. Thus, $\Sigma_1$ and $\Sigma_2$ are isomorphic.

Now let $R$ and $\bar{R}$ be the controllability matrices of $\Sigma_1$ and $\Sigma_2$ respectively and $W$ and $\bar{W}$ be their observability matrices. Then I've already shown that $\bar{R}=SR$ and $\bar{W}=WS^{-1}$. Now I need to show that system $\Sigma_1$ is observable iff $\Sigma_2$ is observable and similarly for controllable.

Now if $\Sigma_1$ is observable, the matrix $$R = (B\ \ AB\ \ A^2B\ \ \ldots\ \ A^{n-1}B)$$ has rank $n$, where $n$ is the dimension of the square matrix $A$. So I have to show that $$\bar{R} = (SB\ \ SAB\ \ SA^2B\ \ \ldots\ \ SA^{n-1}B)=SR$$ also has rank $n$. Similarly for the observability matrix. So basically what I need to show is that a matrix $R$ of full rank keeps full rank when multiplied by a non singular matrix $S$ of appropriate dimensions. This is probably due to the non singularity of $S$ but I am not sure how to show this.

Then finally I have to show that if $\Sigma_1$ and $\Sigma_2$ are observable then $S$ is unique.

Thanks for any help, I tagged this question as linear algebra as well because I think that methods from linear algebra may be used to fairly easily solve the problem.


Consider this:

Suppose $\text{rank} R = n$ but $\text{rank} SR < n$. Then there exists a non-zero vector $x$ such that $SRx = 0$ which implies that $Rx = 0$ also, because $S$ is non-singular (just multiply both sides with $S^{-1}$). But this is impossible since $R$ is non-singular. We obtained a contradiction, hence $\text{rank} SR = n$. The necessity part can also be shown similarly. Also, observability proof follows similarly.

However, we can find infinitely many non-singular matrices $S$, even if the system is not controllable and/or observable. What this theorem says is, controllability/observability is invariant under similarity transformations.

  • $\begingroup$ Thanks a lot for your answer that was very clear. So you are saying that $S$ will in fact not be unique. This makes sense, Thanks again! $\endgroup$ – Slugger Nov 27 '13 at 16:27

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