First of all, I would like to link this question to another one about the inverse of a state space representation:
Inverse of State-space representation (control)
I understand the prove as given on the previous page, but I want to prove it another way. I want to show that $G(s)G^{-1}(s)=I$. However, I get stuck when trying to prove this:
When G(s) is given as: $G(s) = \begin{bmatrix}A&B\\C&D\end{bmatrix}$
Then the inverse is given as: $G^{-1}(s) = \begin{bmatrix}A-BD^{-1}C &BD^{-1}\\-D^{-1}C&D^{-1}\end{bmatrix}$
From these state space representations, we get:
$G(s)G^{-1}(s) = [C(sI-A)^{-1}B+D][-D^{-1}C(sI-A+BD^{-1}C)^{-1}BD{-1}+D^{-1}]$
$= -C(sI-A)^{-1}BD^{-1}C(sI-A+BD^{-1}C)^{-1}BD^{-1} + C(sI-A)^{-1}BD^{-1} + C(sI-A+BD^{-1}C)^{-1}BD^{-1} + I $
Here is where I get stuck. Since $DD^{-1}=I$, the rest should equal the zero matrix, but I fail to prove this.
Anyone who can help me out?
You have the sign of a term wrong. Other than that you can write $$-C(sI-A)^{-1}BD^{-1}C(sI-A+BD^{-1}C)^{-1} + C(sI-A)^{-1} - C(sI-A+BD^{-1}C)^{-1}=\left[-C(sI-A)^{-1}BD^{-1}C+C(sI-A)^{-1}(sI-A+BD^{-1}C)-C\right](sI-A+BD^{-1}C)^{-1}=\left[-C(sI-A)^{-1}BD^{-1}C+C(sI-A)^{-1}(sI-A)+C(sI-A)^{-1}BD^{-1}C-C\right](sI-A+BD^{-1}C)^{-1}=0$$