Converting state-space equations to parallel form Im givin the state space equations shown below and been asked to convert them into parallel form.
$$\begin{bmatrix} \dot z_1(t) \\ \dot z_2(t) \end{bmatrix} = \begin{bmatrix}-4 & -6\\3 & 5\end{bmatrix} \cdot \begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix} + \begin{bmatrix}-2 & -3\\2 & 2\end{bmatrix} \cdot \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}$$
$$\begin{bmatrix} y(t) \end{bmatrix} = \begin{bmatrix}-4 & -6\end{bmatrix} \cdot \begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix}$$
I understand how to get the diagonalized matrix using eigenvectors $T = \begin{bmatrix} v_1 & v_2 \end{bmatrix}$  and $\hat A = T^-1 A T $ but I'm unsure why matrices B and C change.
$$\begin{bmatrix} \dot z_1(t) \\ \dot z_2(t) \end{bmatrix} = \begin{bmatrix}2 & 0\\0 & -1\end{bmatrix} \cdot \begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix} + \begin{bmatrix}2\sqrt{2} & \sqrt{2}\\0 & \sqrt{5}\end{bmatrix} \cdot \begin{bmatrix} x_1(t) \\ x_2(t) \end{bmatrix}$$
$$\begin{bmatrix} y(t) \end{bmatrix} = \begin{bmatrix}1/\sqrt{2} & -1/\sqrt{5}\end{bmatrix} \cdot \begin{bmatrix} z_1(t) \\ z_2(t) \end{bmatrix}$$
can anyone help explain to me what is happenening to matrices B and C?

You have
$ \mathbf{\dot{z}} = A \mathbf{z} + B \mathbf{x} $
$ y = \mathbf{c}^T \mathbf{z} $
With, $A = \begin{bmatrix} -4 && -6 \\ 3 && 5 \end{bmatrix}, \ \ B = \begin{bmatrix} -2 && - 3 \\ 2 && 2 \end{bmatrix}, \ \ \mathbf{c} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} $
To diagonalize $A$ we find the characteristic polynomial $\det(\lambda I - A) $
$\det(\lambda I - A ) = (\lambda + 4)(\lambda - 5) + 18 = \lambda^2 - \lambda -2 = (\lambda - 2)(\lambda + 1)$
Therefore, the eigenvalues are $\lambda_1 = 2 , \lambda_2 = -1 $
The corresponding normalized (upto a sign) eigenvectors (details omitted) are
$ \mathbf{v_1} = \begin{bmatrix} -\dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \end{bmatrix}, \ \ \mathbf{v_2} = \begin{bmatrix} -\dfrac{2}{\sqrt{5}} \\ \dfrac{1}{\sqrt{5}} \end{bmatrix} $
Define $P = [ \mathbf{v_1} , \ \mathbf{v_2} ] $, and $D = \text{diag}(2, -1) $ then
$ A = P D P^{-1} $
Now define a change of variable $\mathbf{w}$ such that, $ \mathbf{z} = P \mathbf{w} $, then it follows that
$ P \mathbf{\dot{w}} = A P \mathbf{w} + B \mathbf{x} $
so that
$\mathbf{\dot{w}} = P^{-1} A P \mathbf{w} + P^{-1} B \mathbf{x} $
Since $A = P D P^{-1}$ , then $ P^{-1} A P = D $
hence,
$\mathbf{\dot{w}} = D \mathbf{w} + P^{-1} B \mathbf{x} $
In addition, the output equation becomes,
$ y = \mathbf{c}^T P \mathbf{w} $
Now
$P^{-1} = \begin{bmatrix} \sqrt{2} && 2 \sqrt{2} \\ - \sqrt{5} && - \sqrt{5} \end{bmatrix} $
Hence,
$P^{-1} B =\begin{bmatrix} \sqrt{2} && 2 \sqrt{2} \\ - \sqrt{5} && - \sqrt{5} \end{bmatrix} \begin{bmatrix} -2 && - 3 \\ 2 && 2 \end{bmatrix} = \begin{bmatrix} 2 \sqrt{2} && \sqrt{2} \\ 0 && \sqrt{5} \end{bmatrix} $
And
$ \mathbf{c}^T P = \begin{bmatrix} 2 && 3 \end{bmatrix} \begin{bmatrix}  -\dfrac{1}{\sqrt{2}} && -\dfrac{2}{\sqrt{5}} \\ \dfrac{1}{\sqrt{2}} && \dfrac{1}{\sqrt{5}} \end{bmatrix} =\begin{bmatrix} \dfrac{1}{ \sqrt{2}} && -\dfrac{1}{\sqrt{5}} \end{bmatrix} $
