A question about reversibility of matrix

Question

Assume that $A$ and $B$ are different $n\times n$ matrices with $A^3＝B^3, AB^2＝B^2A.$ The question is whether $A^2＋B^2$ is invertible or not.

I tried $$(A^2＋B^2)(A-B)＝A^3-B^3＋B^2A-A^2B＝B^2A-A^2B$$,but the assumption is $B^2A＝AB^2$, so I don't know how to do next.

Hope for your help.Thanks.

Answer 1

We assume that $A,B\in M_n(\mathbb{C}),A^3=B^3,AB^2=B^2A$.

$\textbf{Proposition}$.When $B$ is invertible (then $A$ too), $A,B$ are simultaneously triangularizable.

$\textbf{Proof}$. Since $AB^2=B^2A$, $A,B^2$ are simultaneously triangularizable: $A=PUP^{-1},B^2=PVP^{-1}$ where $U,V$ are upper-triangular. Then $B=PU^3V^{-1}P^{-1}$, where $U^3V^{-1}$ is triangular. $\square$

In the sequel, we assume that $A,B$ are invertible.

According to the above proposition, if $spectrum(A)=(\lambda_j)_j$, then there is an ordering $(\mu_j)_j$ of $spectrum(B)$ s.t. ${\lambda_j}^3={\mu_j}^3$ and $\det(A^2+B^2)=\Pi_j({\lambda_j}^2+{\mu_j}^2)$; we conclude that $A^2+B^2$ is invertible.

Answer 2

Let $A=0$, $B$ be nilpotent of index $2$, that is, $B^2=0, B\neq0$. Then $A\neq B$, $A^3=0=B^3$ and $AB^2=0=B^2A$, but $A^2+B^2=0$, which is not invertible.