I have three input/state/output representations of the form
$$\begin{cases}\frac{d}{dt}x=Ax+Bu \\ y=Cx\end{cases}$$
with the three systems given by:
$$A_{1}=\begin{bmatrix} -1 & 0 \\ 0 & -2 \end{bmatrix} B_{1}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} C_{1}=\begin{bmatrix} 1 & 0 \end{bmatrix}$$
$$A_{2}=\begin{bmatrix} -2 & 0 \\ 0 & -1 \end{bmatrix} B_{2}=\begin{bmatrix} 0 \\ 1 \end{bmatrix} C_{2}=\begin{bmatrix} 0 & 1 \end{bmatrix}$$
$$A_{3}=\begin{bmatrix} -1 & 0 \\ 0 & 0 \end{bmatrix} B_{3}=\begin{bmatrix} 1 \\ 0 \end{bmatrix} C_{3}=\begin{bmatrix} 1 & 0 \end{bmatrix}$$
I want to show that these three systems define the same i/o behaviour.
There is a theorem which says that two state space representations are input/output equivalent (i.e. define the same behaviour) if $\exists$ a non-singular matrix $P\in\mathbb{R}^{n\times n}$ with $PA_{1}P^{-1}=A_{2}$, $PB_{1}=B_{2}$ and $C_{1}P^{-1}=C_{2}$.
I took $P=\begin{bmatrix}0 & 1 \\ 1 & 0\end{bmatrix}$ and then, after performing the computations I found that the first system defined the same behaviour as the second system.
Now, I would appreciate some help with showing this is true for all three systems.
I think I can write the i/s/o representation in the form
$$\frac{d}{dt}\frac{1}{y}=\frac{AC}{y}+Bu$$
Since $x=\frac{C}{y}\implies\frac{d}{dt}x=\frac{d}{dt}\frac{1}{y}$
I'm not sure where to go from here.
The third system does not have the same behavior with the first two. But they have the same input/output behavior, i.e. they have the same transfer function. You can calculate the transfer function as
$$ G(s) = C (sI - A)^{-1} B $$
which will give the same result for the three systems above. It is due to the fact that the eigenvalue -2 in the first two systems and 0 in the third system is not controllable (it is also not observable but only one of them is enough for not to see it in the transfer function).