State Space to Diagonal Form (via Eigenvectors) I have a state space representation of a system:
$$\dot x=Ax+Bu$$
$$y=Cx+Du$$
I need to get it into diagonal (decoupled) form. I know if I find the eigenvectors of A (call them P), and do the operation $P^{-1} A P$ I will get a matrix with the eigenvalues of A along the main diagonal.
I'm new to linear algebra, so I'm struggling on how to get the "P" in between the A and x ($APx$).
I read online that if I defined $z=Px$,
$$\dot z=Az+B^{*}u$$ 
$$y=C^{*}z+D^{*}u$$ 
then I can do the following:
$$\dot z=P \dot x=APx+B^{*}u$$ 
$$P^{-1}P \dot x=\dot x=P^{-1}APx+P^{-1}B^{*}u$$ 
$$ y=C^{*}Px+D^{*}u$$
Where I'm stuck is how to figure out what B*, y*, C*, and D* are. My first guess was the following:
$$B^{*}=BP$$
$$C^{*}=P^{-1}C$$
$$D^{*}=DP$$
But that was all based on intuition, and I'm not sure how to check it's validity.
 A: I think the easiest way to understand this is by realizing there are two different parts to this, a change of variables and one matrix multiplication.
First, you do a change of variables $Pz = x$, so $P\dot z = \dot x$, you insert this into your equations, yielding:
$$\begin{align}
\underbrace{P\dot z}_{= \dot x} &= A\underbrace{Pz}_{= x} + Bu \\
y &= CPz + Du
\end{align}$$
and you can note that this does not actually change the equation. You can then multiply with $P^{-1}$ on the left in your first equation:
$$\begin{align}
P^{-1}P\dot z &= P^{-1}APz + P^{-1}Bu \\
y &= CPz + Du
\end{align}$$
and if you simplify this you get:
$$\begin{align}
\dot z &= Dz + P^{-1}Bu \\
y &= CPz + Du
\end{align}$$
which has a diagonal state matrix $D$, using different states, but with the same output. If you want to recover your original states $x$ for some reason, just remember that $x = Pz$.
Of course, you can rename your matrices if you want:
$$\begin{align}
\dot z &= Dz + B_2u \\
y &= C_2z + Du
\end{align}$$
where $B_2 = P^{-1}B$ and $C_2 = CP$.
