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like this : Given $$A,B\in M_{3\times 3}(R) $$and $$A=\begin{pmatrix}1&-3&0\\3&4&-3\\3&3&-2\end{pmatrix} $$ $$B=\begin{pmatrix}1&0&0\\0&1&3\\0&-3&1\end{pmatrix} $$ and$$A=P^{-1}BP$$ How to find the P?

I try to find the Jordan form of A,B ,but $$det(A-\lambda I)=-(\lambda-1)(\lambda^2-2\lambda+10)$$ but $$A,B\in M_{3\times 3}(R) $$ $$\lambda\in R $$ can't find the Jordan form?

or is there another process?

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    We usually use this kind of action to diagonalize a matrix, $QAQ^{-1} = D$ where D is diagonal, Presuming both A and B are diagonalizable, if we diagonalize B, then $RBR^{-1} = D$ We are told the matrices are similar so we should get to the same D, I believe (worth working out). So my thought is $ A = Q^{-1}DQ = Q^{-1}RBR^{-1}Q = P^{1}B P$ where $P = R^{-1} Q$ We know how to find D and Q and R with eigenvalues and eigenvectors so this is where I would start. If the matrices are not diagonalizable then that would be another issue. – 2017-02-09

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P is not unique because if it is multiplied by any nonsingular diagonal matrix, $P^{-1}BP$ is the same.

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Multiplying both sides by $P$ on the left yields $PA=BP$ or $-BP+PA=C$ where $C$ is all-zero. This is called a Sylvester equation.

The equation has a unique solution for $P$ when there are no common eigenvalues of $A$ and $B$.

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    It looks like you accidentally swapped P and A in the second equation. – 2017-02-09