State transform from one state space representation to another I have a state space representation, system S1, in the form of:
$$ \frac{dx}{dt} = Ax + bu $$ $$y = c^Tx$$
This system is transformed with the state transform $$x=T z$$
into the system S2:
$$ \frac{dz}{dt} = \begin{pmatrix} -1 & -2  \\-1 & -2  \end{pmatrix}z + \begin{pmatrix} 0  \\1  \end{pmatrix}u $$ $$y = \begin{pmatrix} 1  &-1  \end{pmatrix}z $$
T is the transformation matrix.
The only thing I know from system S1 is, that is in diagonal form.
So $A$ should look something like this: A = $\begin{pmatrix} a & 0  \\0 & d  \end{pmatrix}$
I think I know how I could transform S1 into S2 but I don't know the other way.
Found some formulas like $$T^{-1}AT = \begin{pmatrix} -1 & -2  \\-1 & -2  \end{pmatrix}$$
and $$T^{-1}b = \begin{pmatrix} 0  \\1  \end{pmatrix}$$
I have a problem obtaining the transformation matrix.
I also thought about the eigenvalues of S2 (0, -3), but I don't know if I am right that these values are the diagonal points of the matrix A.
edit:
So with @Sasha 's help I got the system matrix A:
$$p_1 = \begin{pmatrix} -2  \\ 1  \end{pmatrix}$$
$$p_2 = \begin{pmatrix} 1  \\ 1  \end{pmatrix}$$
$$T = \begin{pmatrix} -2 & 1  \\1 & 1  \end{pmatrix}$$
$$T^{-1}\begin{pmatrix} -1 & -2  \\-1 & -2  \end{pmatrix}T = \begin{pmatrix} 0 & 0  \\0 & -3  \end{pmatrix}$$
$$b = \begin{pmatrix} 1  \\ 1  \end{pmatrix}$$
$$c = \begin{pmatrix} -\frac{2}{3}  \\ -\frac{1}{3}  \end{pmatrix}$$
Some additional infos:
Matrix A
Matrix A is the system matrix, and relates how the current state affects the state change x' . If the state change is not dependent on the current state, A will be the zero matrix. The exponential of the state matrix, eAt is called the state transition matrix.
Matrix B
Matrix B is the control matrix, and determines how the system input affects the state change. If the state change is not dependent on the system input, then B will be the zero matrix.
Matrix C
Matrix C is the output matrix, and determines the relationship between the system state and the system output.
 A: Suppose an invertible transformation matrix is given such that the states of the transformed system becomes $q$ i.e. $q=Tx$. Then, we also know that $\dot q = T\dot x$. From these, we obtain $T^{-1}q=x$ and $T^{-1}\dot q=\dot x$. If you plug these in to the state space representation:
$$\begin{align}
T^{-1}\dot q&= AT^{-1}q+Bu\\
y &= CT^{-1}q + Du
\end{align}
$$
Now multiply the first equation with $T$,
$$\begin{align}
\dot q&= TAT^{-1}q+TBu\\
y &= CT^{-1}q + Du
\end{align}
$$
This is the same system $G(s)$ in different state coordinates. In your particular case, things are slightly easier since you are looking for the transformation matrices that diagonalizes $A$ which is possible, for example, with eigenvalue decomposition.
A: Transformation formulas you wrote are correct. Because $A$ is diagonal, you can tell that $a$ and $d$ must eigenvalues. And the matrix $T$ can be constructed from eigenvectors of $\left( \begin{array}{cc} -1 & -2 \\ -1 & -2 \end{array} \right)$, so that columns of $T$ will be normalized eigenvectors.
