The following lemma is stated without proof in the lectures notes on Model Reduction by M. Voigt (available here)
Lemma 4.4 Let $[A,B,C,D]\in\Sigma_{n,m,p}$ be asymptotically stable. Let $T\in\mathbb{R}^{n\times n}$ be invertible and define $[\tilde A,\tilde B, \tilde C,\tilde D]:=[T^{-1}AT,T^{-1}B,CT,D]$. Then,
$P$ is the controllability Gramian of $[A,B,C,D]$, if and only if $\tilde P:=T^{-1}PT^{-T}$ is the controllability Gramian of $[\tilde A,\tilde B, \tilde C,\tilde D]$.
$Q$ is the observability Gramian of $[A,B,C,D]$, if and only if $\tilde Q:=T^{T}QT^{-1}$ is the observability Gramian of $[\tilde A,\tilde B, \tilde C,\tilde D]$.
I am not sure on how to prove this result. Would anyone has some hints? Thanks!
My idea for the 1. question;
Using the Lyapunov equations, $$\tilde{A} \tilde{P} + \tilde{P} \tilde{A}^T = -\tilde{B}\tilde{B}^T \hspace{1.6cm} (1)$$ $$A P + P A^T = -B B^T \hspace{1.6cm} (2)$$ $$\tilde{A} = T^{-1} A T; \hspace{0.6cm} \tilde{B} = T^{-1}B \hspace{1.6cm} (3)$$ Then inserting (3) into (1) we have, $$T^{-1} A T \tilde{P} + \tilde{P} {(T^{-1} A T)}^T = -T^{-1}B {(T^{-1}B)}^T$$ $$T^{-1} A T \tilde{P} + \tilde{P} T^T A^T T^{-T} = -T^{-1}B B^T T^{-T}$$
Now we use (2) on the right side and obtain, $$ T^{-1} A T \tilde{P} + \tilde{P} T^T A^T T^{-T} = T^{-1} (A P + P A^T ) T^{-T}$$ By multiplying the whole equation with $T$ on the left and $T^T$ on the right we have, $$ A T \tilde{P} T^T + T \tilde{P} T^T A^T = A P + P A^T$$ If we take a substitution $M = T \tilde{P} T^T$, $$ A M + M A^T = A P + P A^T$$ We clearly see that this is, $$M = P = T \tilde{P} T^T$$ Finally, by multiplying the whole equation with $T^{-1}$ on the left and $T^{-T}$ on the right we obtain, $$ T^{-1} P T^{-T} = \tilde{P}$$