Proving that similarity transformation of state-space preserves the Euclidean norm I am aware that the state space realization of a dynamical system is not unique. So if we have a dynamical system:
$\dot{x} = Ax + Bu$
$y = Cx$
Then we can write it as
$\begin{bmatrix}
\dot{x}\\y
\end{bmatrix}
= G
\begin{bmatrix}x\\u\end{bmatrix}
$
where G is
$G = \begin{bmatrix}
A \ \ \ \ | & B\\ 
\hline
C \ \ \ \ | & 0
\end{bmatrix}
$
but one can also formulate it as
$G = \begin{bmatrix}
TAT^{-1} \ \ | & TB\\ 
\hline
CT^{-1} \ \  \ \ \ | & 0
\end{bmatrix}$
for any invertible matrix $T$. 
I need to show that these two realizations of $G$ have the same Euclidean norm. I have found a video showing how similarity transformations preserve the trace and the determinant, but they did just an example problem, without doing the proof.
Edit:
It has been given that $G \in \mathcal{RH}_2$, where $\mathcal{RH}_2$ is a set of all real rational, strictly proper and stable transfer matrices. It is a Hilbert Space with the inner product:
$\langle F,G \rangle = \sup_{\sigma > 0} \left\{ \frac{1}{2\pi}\int_{-\infty}^{\infty} trace \left\{ F^* (\sigma+j\omega) G(\sigma+j\omega) \right\} d\omega \right\}$
and the corresponding norm given by
${{\left\Vert F \right\Vert}_{2}}^2 = \frac{1}{2\pi} \int_{-\infty}^{\infty} trace \left\{ 
F^*(j\omega)F(j\omega) \right\}d\omega$
 A: The problem here is really the notation.
This is adding to the answer by obareey a bit more details. Given a linear system 
$$
\begin{align}
\dot{x} &= A x + B u \\
y &= C x + D u \\
\end{align}
$$
In this question $D = 0$, but that doesn't really change anything. Then i.e. this is wrong:
$$
G = \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$
As mentioned by obareey, $G$ is not a real matrix but instead a transfer function matrix. Often this is written instead like
$$
G(s) \sim \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$
sometimes also written as 
$$
G(s) \triangleq \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right] \text{ or } G(s) \overset{s}= \left[  \begin{array}{c|c} A&B\\ \hline C&D \end{array} \right]
$$
Unfortunatly the notation varies and I have seen all of these. Some authors even use an equal sign, which makes things really confusing (or: wrong). However they really all mean the same, namely that $G(s) = C(s I - A)^{-1} B + D$ and $I$ is an identity matrix. 
The $2$-norm of a general MIMO transfer function is
$$
\Vert G(s) \Vert_2 = \Big( \frac{1}{2 \pi} \int_{-\infty}^{\infty} \text{trace} \big[ G(j \omega)^H G(j \omega) \big] d \omega \Big)^{1/2}
$$
so of course $\Vert G(s) \Vert_2 = \Vert \widetilde{G}(s) \Vert_2$ when $G(s) = \widetilde{G}(s)$, assuming that $A$ is stable so that the integral exists, i.e. $\Vert G(s) \Vert_2 < \infty$. So all you need to show is that
$$
C(s I - A)^{-1} B + D = \widetilde{C}(s I - \widetilde{A})^{-1} \widetilde{B} + \widetilde{D}
$$
using an invertible $T$ and
$$
\begin{align}
\widetilde{A} &= T A T^{-1} \\
\widetilde{B} &= T B \\
\widetilde{C} &= C T^{-1} \\
\widetilde{D} &= D
\end{align}
$$
This is standard realization theory. You can check that both transfer functions are the same:
$$
\begin{align}
\widetilde{G}(s) &= \widetilde{C}(s I - \widetilde{A})^{-1} \widetilde{B} + \widetilde{D} \\
&= C T^{-1}(s I - T A T^{-1})^{-1} T B + D \\
&= C \big(T^{-1}(s I - T A T^{-1})T\big)^{-1}B + D \\
&= C \big(T^{-1} s I T - T^{-1} T A T^{-1} T\big)^{-1}B + D \\
&= C(s I - A)^{-1} B + D \\
&= G(s)
\end{align}
$$
because $T^{-1} s I T = s(T^{-1} I T)  = s(T^{-1} T) = s I$ since $T^{-1} T = I$ and the fact that $(K_1 K_2 K_3)^{-1} = K_3^{-1} K_2^{-1} K_1^{-1}$ for some invertible matrices $K_1, K_2, K_3$.
So because $G(s) = \widetilde{G}(s)$ their $2$-norms are the same.
A: In this context $G$ is not a matrix, but a representation of the system. The 2-norm is not a matrix norm, but a system norm. This 2-norm is still defined by the system's frequency response. Since the transfer function matrix (hence the frequency response) of the system is invariant under similarity transformations, its 2-norm also does not change.
A: It is not the case that the transformed matrix $G$ will have the same $2$-norm. For instance, consider
$$
A = \pmatrix{0&1\\0&0}, \quad B = \pmatrix{1\\0}, \quad C = \pmatrix{1&0}, \quad T = \pmatrix{2&0\\0&1}.
$$
We find that the matrices in question are
$$
G = \left[ 
\begin{array}{cc|c}
0&1&1\\
0&0&0\\
\hline
1&0&0
\end{array}
\right], \quad 
G' = \left[ 
\begin{array}{cc|c}
0&2&2\\
0&0&0\\
\hline
\frac 12&0&0
\end{array}
\right].
$$
Verify that these two matrices do not have the same $2$-norm.

For an example of a stable system for which this occurs, it suffices to replace $A$ with the matrix
$$
\pmatrix{-1&1\\0&-1}
$$
in the above example.
