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I want to show that

$(I-H^T(-s)H(s))^{-1}$ has no poles on the imaginary axis

with $H(s)=C(sI-A)^{-1}B$ and $H^T(-s)=-B^T(sI+A)^{-T}C^T$

is equivalent to $M_\gamma$ has no purely imaginary eigenvalues.

$M_\gamma = \left( \begin{matrix} A & \frac{1}{\gamma}BB^{T} \\ -\frac{1}{\gamma}C^{T}C & -A^T \\ \end{matrix} \right) $

My approach: $n=dim(A)$

$det(sI-M_\gamma)=det(sI-A)det(sI+A^T)+\frac{1}{\gamma^{2n}}det(BB^{T})det(C^{T}C)=0$ $\implies 1+det(\frac{1}{\gamma^2}BB^T(sI+A)^{-T}C^TC(sI-A)^{-1})=0$

But it is still not:

$I+\frac{1}{\gamma^2}B^T(sI+A)^{-T}C^TC(sI-A)^{-1}B=(I-H^T(-s)H(s))=0$

Where is my mistake? Sign mistake?

Remark: $H(s)$ is a strictly causal and stable state-space system and s is the Laplace-variable.

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    $\begingroup$ I'm voting to close this question as off-topic because it sounds like it has been assigned as an exercise $\endgroup$
    – Yemon Choi
    Commented May 19, 2016 at 17:04
  • $\begingroup$ No, it is not an exercise. I want to understand the proof: ocw.mit.edu/courses/electrical-engineering-and-computer-science/… $\endgroup$
    – Sebastian
    Commented May 19, 2016 at 17:13
  • $\begingroup$ It would help much more if you would give me some hints on my mistakes instead of putting it as an off-topic. $\endgroup$
    – Sebastian
    Commented May 20, 2016 at 6:41
  • $\begingroup$ I think math.stackexchange.com might be a better site for this question $\endgroup$
    – Yemon Choi
    Commented May 20, 2016 at 10:16

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