Proving that similarity transformation of state-space preserves the Euclidean norm

Question

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

${{\left\Vert F \right\Vert}_{2}}^2 = \frac{1}{2\pi} \int_{-\infty}^{\infty} trace \left\{ F^*(j\omega)F(j\omega) \right\}d\omega$

Answer 1

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.

Answer 2

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.

Answer 3

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.

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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.