$A,B$ matrices $n \times n$ have the same eigenvectors. $M_A=(x+1)^2, P_B=x^5.$ Prove $B^3=0.$

Question

$A,B$ matrices $n \times n$ have the same eigenvectors.

$M_A=(x+1)^2$ (minimal polynomial), $P_B=x^5$ (characteristic polynomial).

Prove $B^3=0.$

My solution:

Let $J_A$ a jordan matrix which similar to $A , J_B$ a jordan matrix which similar to $B$.

The numbers of jordan blocks in jordan matrix determined by the number of the eigenvectors, since $A,B$ have the same eigenvectors then $J_A,J_B$ has the same form.

$M_A=(x+1)^2 \implies$ the largest Jordan block of $J_A$ (and $J_B$) is size $2 \times 2$.

$P_B = x^5 \implies n=5 \implies J_B (\text{and } J_A) \text{ are } 5\times5 \text{ matrices}.$

$J_B=\left\{ J_{2}(0) , J_{2}(0) , J_{1}(0)\right\}$ or $\left\{ J_{2}(0) , J_{1}(0) , J_{1}(0),J_{1}(0) \right\}.$

There is invertible matrix $P$ such that $P^{-1}BP=J_{B}.$

$(P^{-1}BP)^3=P^{-1}B^3P=J_{B}^3=\left\{ (J_{2}(0))^3 , (J_{2}(0))^3 , (J_{1}(0))^3\right\} = \left\{ (J_{2}(0))^3 , (J_{1}(0))^3 , (J_{1}(0))^3,(J_{1}(0))^3 \right\} = 0_{5 \times 5}$

Is my proof correct ?

I'd be grateful for your help!

Answer 1

In order to remove this question from the unanswered queue, let me say that your proof seems to be right, and moreover, we get $B^2=0$.