Solve this "quasi diagonalization" matrix equation I would like to know if it is possible to solve for matrix T any matrix equation in the form of 
T^-1 * F * T = G
where F and G are nxn known matrix.
How can I solve equation and find such matrix T?
In my example, with Matlab representation of matrices 
F = [0 -1 0; 2 1 0; 0 1 0]
and 
G = [0 2 0; -1 1 0; -2 2 0]
I see it resemble the diagonalization form problem, however G in this case is not diagonal as you see.
By an ansatz, I checked with Matlab and confirmed my guess that F and G have the same eigenvalues, but I cannot go much far. 
I am studying Kalman observability matrix canonical form in state space control system (It is used to separate observable and unobservable states)
I  want to understand if my solution (matrix F) is similar to the solution provided by the book ( matrix G ) by any change of basis, that will prove
such Kalman matrix are not unique. 
ALso, am I right in such last quote?
Thanks!
 A: This can be written as a Sylvester Equation
$$FX-XG=0$$
A direct approach using vectorization (by column stacking) yields
$$\eqalign{
x &= \operatorname{vec}(X) \\
(I\otimes F)x &= (G^T\otimes I)x \\
Ax &= Bx \\
}$$
which is the generalized eigenvalue equation $(Ax=\lambda Bx)$ with $\lambda=1$. 
In your example, both $(F,G)$ are singular, and therefore $(A,B)$ are also singular. So the problem cannot be transformed into a standard eigenvalue equation by multiplying by a matrix inverse.
However, both Matlab and Julia have an eigen function which can calculate the eigenvalues/vectors for such equations.
A solution is not guaranteed, but if $\,{\tt1}$ occurs as an eigenvalue, then any associated eigenvectors are solutions. 
For your example, Julia found two suitable eigenpairs:
$$\eqalign{
\lambda_1 = 1, 
\quad 
x_1 &= \frac{1}{8}\left(\begin{array}{r}
-3\\-2\\3\\-2\\8\\2\\0\\0\\0
\end{array}\right), 
\quad
X_1 = \operatorname{Mat}(x_1)
  = \frac{1}{8}\left(\begin{array}{r}
-3&-2&0\\-2&8&0\\3&2&0
\end{array}\right)
\\
\lambda_2 = 1, 
\quad
x_2 &= \left(\begin{array}{r}
0.883634\\-0.767268\\-0.883634\\-0.767268\\-1.000000\\0.767268\\0.000000\\0.000000\\0.000000
\end{array}\right), 
\quad
X_2 = \left(\begin{array}{r}
0.883634&-0.767268&0\\-0.767268&-1.000000&0\\-0.883634&0.767268&0
\end{array}\right) \\
\\
}$$
Interestingly both matrices are of the form
$$
\left(\begin{array}{rcr}
-\alpha&-\beta&0\\-\beta&\operatorname{sign}(\alpha)&0\\ \alpha&\beta&0
\end{array}\right)
$$
