In http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/lecture-notes/MIT6_241JS11_lec12.pdf on pages 11-12 it is said:
For a stable and strictly causal state-space system $H(s)=C(sI-A)^{-1}B$ the following statements are equivalent:
- $(I-\frac{1}{\gamma^2}H^T(-s)H(s))^{-1}$ has no poles on the imaginary axis (with $H^T(-s)=-B^T(sI+A)^{-T}C^T$).
- $M_\gamma = \left( \begin{matrix} A & \frac{1}{\gamma}BB^{T} \\ -\frac{1}{\gamma}C^{T}C & -A^T \\ \end{matrix} \right) $ has no purely imaginary eigenvalues.
But how can I show it? It seems to be a simple matrix transformation, but I have no idea on how to do it.
Something like:
$det(sI-M_\gamma)=det(sI-A)det(sI+A^T)+det(\frac{1}{\gamma^{2}}BB^{T})det(C^{T}C)=0$ $\implies 1+det(\frac{1}{\gamma^2}BB^T(sI+A)^{-T}C^TC(sI-A)^{-1})=0$
And then I have to reach somehow:
$I+\frac{1}{\gamma^2}B^T(sI+A)^{-T}C^TC(sI-A)^{-1}B=(I-H^T(-s)H(s))=0$
Note: I am not a student there. But I try to understand the mathematics, because it is necessary for another problem.
I have the feeling that it might be $(I-\frac{1}{\gamma^2}H(s)H^T(-s))^{-1}$ and not $(I-\frac{1}{\gamma^2}H^T(-s)H(s))^{-1}$. Because the Matlab commands
A=[1 2;3 4];B=[3;0];C=[2 0];D=0;H=ss(A,B,C,D);HT=ss(-A',-C',B',0);(1+HT*H)^-1 results in something that looks like the eigenvalues of $ \left( \begin{matrix} -A^T & -\frac{1}{\gamma}C^{T}C \\ -\frac{1}{\gamma}BB^{T} & A \\ \end{matrix} \right) $.