How can I prove that $A$ and $B$ are not similar matrices? 
Prove that $A$ and $B$ have the same characteristic polynomial, but A
and B are not similar \begin{align*} A=\begin{pmatrix} 0 & 1 & 0\\  0
 & 0 &1 \\ 
 -2 & 3 & 0 \end{pmatrix} \ \ \ \ \ \ \ \ \ B=\begin{pmatrix} 1 & 0 & 0\\  0 & 1 &0 \\  0 & 0 & -2 \end{pmatrix} \end{align*}

I've already proved that: $det(A-tI)=-t^3+3t-2=det(B-tI)$, so A and B have the same characteristic polynomial.
I'm not sure how can I prove that A and B are not similar. I know that if they were similar then it means that there exists a invertible matrix such that $A=M^{-1}BM$. Any idea to prove that $A$ and $B$ are not similar?

Suppose they are similar.
Then that means $A$ is diagonalizable with eigenvalues of $(1,1,-2)$. But $A$ is a Companion Matrix which is diagonalizable over $\mathbb C$ iff there all eigenvalues are distinct, which is a contradiction.
