How to find transition matrix for two linear operator matrices I have two matrices which are similar to each other. I need to find transition matrix T: A=$(T)^{-1}$ BT
A=\begin{bmatrix}2&-1&-1\\-3&4&3\\4&-4&-3\end{bmatrix}
and
B=\begin{bmatrix}-1&-1&2\\-1&0&1\\-3&-2&4\end{bmatrix}
I know if matrix are similar then characteristic polynomials are also equal so as  jordan normal form. In this case does it mean that through transition to jordan normal form for each matrix i can find matrix T? Is there any other optimal options to find matrix T?
 A: You can indeed use the fact that the jordan normal forms are the same. Let $S_1$ be the matrix s.t. $S_1^{-1}AS_1 = J$ is in jordan normal form and let $S_2$ be the matrix s.t. $S_2^{-1}BS_2 = J$. Then we have:
$$S_1^{-1}AS_1 = S_2^{-1}BS_2 \Leftrightarrow A= (S_1S_2^{-1}) B (S_2S_1^{-1})$$
which gives $T = S_2S_1^{-1}$.
Another option would be to view $A$ as the matrix describing a linear map with respect to the standard basis $((1,0,0), (0,1,0), (0,0,1))$. Then $B$ is the matrix of that linear map w.r.t. another basis that we need to find in order to obtain the matrix $T$. For that we need to find three vectors $v_1, v_2, v_3$ such that:
$$Av_1 = -v_1 - v_2 - v_3$$
$$Av_2= -v_1 -2v_3$$
$$Av_3 = 2v_1 + v_2 + 4v_3$$
This is a system of nine variables (3 vectors with 3 components) that you could solve. Then $T^{-1}$ would just be the matrix with the columns $(v_1, v_2, v_3)$ and you could get $T$ by inverting this matrix.
I believe that the jordan normal form way of doing this could actually be easier - but it's your decision which route you take.
