Problem about characteristic polynomial, eigenvalues I'm having trouble solving the following question:
Let A be a matrix with the following characteristic polynomial: $\,p(t)=t^3+2t^2-3t\,$. 
Show that the matrix $\,A^2+A-2I\,$ is similar to the matrix$$\begin{pmatrix}0&0&0\\0&4&0\\0&0&\!\!\!-2\end{pmatrix}$$
I've already figured out that the eigenvalues of $A$ are $0,1,-3$, and that $\,A^2+A-2I\,$ equals to $\,(A+2I)(A-I)\,$. Also, the eigenvalues of $\,A+tI\,$ are the eigenvalues of $\,A+t\,$ ($\,+t\,$ to each eigenvalue). The only thing left now is the multiplication of the two matrices, what can I know about it? Is there another way to solve the question? Thanks!
 A: I think I've answered the question:
If $\,x_1,..,x_n\,$ are the eigenvalues of $A$, not only that the eigenvalues of $\,A+tI\,$ are $\,x_1+t,...,x_n+t\,$, but they are also eigenvalues with the same eigenvectors. Similarly $\,A+t_1I\,,\,A+t_2I\,$ have eigenvalues which differ in $\,t_2-t_1\,$ and with the same eigenvectors.
Now, if $\,v_1\,$ is a common eigenvalue of $A$ and $B$, $$\,ABv_1=A(Bv_1)=A(yv_1)=y(Av_1)=y(xv_1)=xyv_1\,$$ Meaning, the eigenvalues of $AB$ are $xy$ where x is $A'$s eigenvalue and $y$ is $B'$s eigenvalue.
So, if the eigen values of $A$ are $0,1,-3$, the eigenvalues of $A+2I$ and $A-I$ are $3,-1,2$ and $0,4,-1$, so the eigenvalues of $(A+2I)(A-I)=3\cdot 0\,,\,4\cdot (-1)$ and $(-1)\cdot 2\,$ , which are $D'$s eigenvalues. The matrices have the same eigenvalues, hence they are similar. Am I correct?
A: Since A has 3 distinct eigenvalues, it is similar to a diagonal matrix:
$$
A=PDP^{-1}
$$
$$
D=\left[\begin{array}{rrr}
1 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & 0 \end{array}\right]
$$
Now your polynomial in A becomes:
$$
A^2+A-2I=(PDP^{-1})^2+(PDP^{-1})-2(PP^{-1})\\=PD^2P^{-1}+PDP^{-1}-2PP^{-1}\\=P(D^2+D-2I)P^{-1}
$$
Hence the matrix $A^2+A-2I$ is similiar to $D^2+D-2I=\left[\begin{array}{rrr}
0 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & -2 \end{array}\right]$
