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?
 A: $B$ is diagonal, hence diagonalizable. To see that $A$ is not diagonalizable, note that $A$ has eigenvalues as the roots of $t^3-3t+2 = (t-1)^2(t+2)$, hence has eigenvalues $1$ and $-2$.
Now, if $Av = v$ then for $v = (v_1,v_2,v_3)$ we get $v_1=v_2,v_2 = v_3$ and $v_3 = 3v_2 - 2v_1$ , so the eigenspace associated to the eigenvalue $1$ is one dimensional and spanned by $(1,1,1)$.  It follows that $A$ is not diagonal, since it has an eigenvalue whose geometric multiplicity is not equal to its algebraic multiplicity.
Finally, if two matrices are similar and one of them is diagonalizable, the other one must be too. It follows that $A$ and $B$ can't be similar.
A: Similar matrices represent the same linear transfromation. Here there the matrix $A$ has only  2 independent eigenvectors $(1,-2,4)^T,(1,1,1)^T$ while the diagonal $B$ has obviously 3 independent eigenvectors. Therefore $A$ must transform a 2-dimensional subspace into a 1-dimesnional one and it cannot correspond to the same linear transformation as $B$.
A: The minimal polynomial of $B$ is $m(t)=(t-1)(t+2)$. However, the $a_{11}$ entry of the following product is non-zero:
$$
       (A-I)(A+2I)=\left[\begin{array}{ccc}-1 & 1 & 0 \\ 0 & -1 & 1 \\-2 & 3 & -1\end{array}\right]\left[\begin{array}{ccc}2 & 1 & 0 \\ 0 & 2 & 1 \\ -2 & 3 & 2\end{array}\right]
$$
A: 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.
