Matrices: knowing that $M^3=N^3$ and that $MN^2=NM^2$, how to prove $M^2+N^2$ is non-invertible?

Question

The exercise: Let $M$ and $N$ be two distinct $n\times n$ matrices such that $M^3=N^3$ and that $MN^2=NM^2$. Prove that $M^2+N^2$ is a non-invertible matrix.

I guess I need to show that the determinant is zero, or create a proof by contradiction, but I can't find a good demonstration.

Answer 1

By contradiction, if it exists $A$ such that $(M^2+N^2)A=I$.

Then

$$M=MI=M(M^2+N^2)A=(M^3+MN^2)A=(N^3+NM^2)A=N(N^2+M^2)A=N$$

In contradiction with the hypothesis $M \neq N$.

Answer 2

Consider $(M-N)(M^2+N^2)=M^3-NM^2+MN^2-N^3$.

Can you finish?

Answer 3

Starting from the identity $MN^2 = NM^2$:

$$ \begin{align*} &\Leftrightarrow MN^2 +M^3= NM^2 +N^3 \\ &\Leftrightarrow M(N^2 +M^2)= N(M^2 +N^2) \\ \end{align*} $$ Now, if $N^2+M^2$ were invertible, we'd arrive at $M=N$.