[proofplan]
The proof is a direct matrix calculation expressing the transformed resolvent through the original resolvent. First we fix the realization dimensions, the similarity matrix, the resolvent sets, and the two transfer functions. We then factor $sI_n-\tilde A$ as $T^{-1}(sI_n-A)T$, which simultaneously proves equality of the two resolvent sets and gives the inverse formula. Substituting the definitions of $\tilde B$, $\tilde C$, and $\tilde D$ cancels the similarity factors and proves equality of the transfer functions.
[/proofplan]
[step:Factor the transformed resolvent through the similarity matrix]
Let $n,m,p \in \mathbb{N}$ be the state, input, and output dimensions of the realization. Thus $A,\tilde A \in \mathbb{C}^{n \times n}$, $B,\tilde B \in \mathbb{C}^{n \times m}$, $C,\tilde C \in \mathbb{C}^{p \times n}$, and $D,\tilde D \in \mathbb{C}^{p \times m}$. Let $T \in \mathbb{C}^{n \times n}$ be the invertible similarity matrix, so $\tilde A=T^{-1}AT$, $\tilde B=T^{-1}B$, $\tilde C=CT$, and $\tilde D=D$. Let $I_n \in \mathbb{C}^{n \times n}$ denote the identity matrix. Define the resolvent sets
\begin{align*}
\rho(A):=\{s \in \mathbb{C}:sI_n-A \text{ is invertible}\},
\qquad
\rho(\tilde A):=\{s \in \mathbb{C}:sI_n-\tilde A \text{ is invertible}\}.
\end{align*}
For $s \in \rho(A)$ and $s \in \rho(\tilde A)$ respectively, define the transfer functions
\begin{align*}
G(s):=C(sI_n-A)^{-1}B+D,
\qquad
\tilde G(s):=\tilde C(sI_n-\tilde A)^{-1}\tilde B+\tilde D.
\end{align*}
Fix $s \in \mathbb{C}$. Since $\tilde A=T^{-1}AT$ and $I_n=T^{-1}I_nT$, we compute
\begin{align*}
sI_n-\tilde A
=
sT^{-1}I_nT-T^{-1}AT.
\end{align*}
Factoring $T^{-1}$ on the left and $T$ on the right gives
\begin{align*}
sI_n-\tilde A
=
T^{-1}(sI_n-A)T.
\end{align*}
Because $T$ is invertible, the matrix $T^{-1}(sI_n-A)T$ is invertible if and only if $sI_n-A$ is invertible. Hence $s \in \rho(\tilde A)$ if and only if $s \in \rho(A)$, so $\rho(\tilde A)=\rho(A)$.
If $s \in \rho(A)$, then the inverse of $sI_n-\tilde A$ is
\begin{align*}
(sI_n-\tilde A)^{-1}
=
T^{-1}(sI_n-A)^{-1}T.
\end{align*}
Indeed, multiplying the proposed inverse on the right gives
\begin{align*}
\bigl(T^{-1}(sI_n-A)T\bigr)\bigl(T^{-1}(sI_n-A)^{-1}T\bigr)
=
T^{-1}(sI_n-A)(sI_n-A)^{-1}T
=
I_n.
\end{align*}
Multiplication in the opposite order gives the same identity, so the displayed formula is the two-sided inverse.
[guided]
Fix $s \in \mathbb{C}$. The point of the similarity transformation is that $\tilde A$ is not an unrelated matrix; it is $A$ written in the new state coordinates determined by $T$. We therefore rewrite $sI_n-\tilde A$ using only $A$ and $T$.
Since $T$ is invertible, we have $I_n=T^{-1}I_nT$. Since $\tilde A=T^{-1}AT$, we get
\begin{align*}
sI_n-\tilde A
=
sT^{-1}I_nT-T^{-1}AT.
\end{align*}
Both terms contain $T^{-1}$ on the left and $T$ on the right, so matrix distributivity gives
\begin{align*}
sI_n-\tilde A
=
T^{-1}(sI_n-A)T.
\end{align*}
This identity also determines exactly when the transformed resolvent exists. Because multiplication by an invertible matrix on the left and right preserves invertibility, $T^{-1}(sI_n-A)T$ is invertible if and only if $sI_n-A$ is invertible. Therefore $s \in \rho(\tilde A)$ if and only if $s \in \rho(A)$, and hence $\rho(\tilde A)=\rho(A)$.
Now assume $s \in \rho(A)$. Then $sI_n-A$ has inverse $(sI_n-A)^{-1}$. The inverse of $T^{-1}(sI_n-A)T$ should undo the right factor $T$, then undo $sI_n-A$, then undo the left factor $T^{-1}$, so the candidate is
\begin{align*}
T^{-1}(sI_n-A)^{-1}T.
\end{align*}
Checking it by multiplication,
\begin{align*}
\bigl(T^{-1}(sI_n-A)T\bigr)\bigl(T^{-1}(sI_n-A)^{-1}T\bigr)
=
T^{-1}(sI_n-A)(sI_n-A)^{-1}T
=
I_n.
\end{align*}
The same multiplication in the reverse order also gives $I_n$. Hence
\begin{align*}
(sI_n-\tilde A)^{-1}
=
T^{-1}(sI_n-A)^{-1}T.
\end{align*}
[/guided]
[/step]
[step:Substitute the transformed realization into the transfer function]
Let $s \in \rho(A)=\rho(\tilde A)$. Using $\tilde C=CT$, $\tilde B=T^{-1}B$, $\tilde D=D$, and the resolvent identity from the previous step, we obtain
\begin{align*}
\tilde C(sI_n-\tilde A)^{-1}\tilde B+\tilde D
=
CT\bigl(T^{-1}(sI_n-A)^{-1}T\bigr)T^{-1}B+D.
\end{align*}
Associativity of matrix multiplication and the identities $TT^{-1}=I_n$ give
\begin{align*}
CT\bigl(T^{-1}(sI_n-A)^{-1}T\bigr)T^{-1}B+D
=
C(sI_n-A)^{-1}B+D.
\end{align*}
Thus $\tilde G(s)=G(s)$ for every $s \in \rho(A)$. This is exactly the asserted transfer-function invariance, and the proof is complete.
[/step]
