State Space invariance to a change of basis

Question

I've been trying to show an identity.

Context:

• $x$,$u$ are vectors
• $A$ has dimension $n\times n$ such that:
• $\dot{x}= Ax+Bu$
• $y= Cx +Du$

We can show that in Laplace domain: $$G(s)= C(sI-A)^{-1}B + D$$ with $G(s)$ given by: $$(Y(s)=G(s)*U(s))$$ That is simple.

The problem is that I can't convince myself that if I use $z=Tx$ we will have the same $G(s)$. So that means: $$G(s)=C(sI-A)^{-1}B + D=CT^{-1}(sI-TAT^{-1})^{-1}TB + D$$ So the problem for me is to show this last identity, I lack of linear algebra knowledge to do that. Thank you.

Answer 1

$$ A' = TAT^{-1}\\ B' = T B\\ C' = C T^{-1}\\ D'=D $$

then

$$ C'(sI-A')^{-1}B'+D' = CT^{-1}(sT T^{-1}+TAT^{-1})^{-1}T B + D = CT^{-1}(T(sI-A)T^{-1})^{-1}TB+D $$

etc.

Answer 2

Well I solved, just used the property that the $T^{-1}M^{-1}T=(T^{-1}MT)^{-1}$. Thank you all.