Let $A\in M_{5×5}(\mathbb{R})$ be a matrix such that $\operatorname{rank}(A)=2$ and $A^3 = 0$. Is A guaranteed to be diagonalizable over R?

Question

Let $A\in M_{5×5}(\mathbb{R})$ be a matrix such that $\operatorname{rank}(A)=2$ and $A^3 = 0$. Is A guaranteed to be diagonalizable over R?

I've been searching my notes for some sort of theorem that would help me with this problem, but I have had no luck.

Without the characteristic polynomial or actual eigenvalues, I can't seem to use the two step test for diagonalizability or any other theorem I know of for that matter.

Any thoughts?

Answer 1

If you are familiar with minimal polynomials, then it is easy to see that the only eigenvalue of $A$ is $0$ and the system $(A-0I)v=0$ has only $3$ linearly independent solutions (eigenvectors) by virtue of $A$ having rank $2$. So $A$ is guaranteed to be non-diagonalizable.

Answer 2

If $A$ was diagonalizable, then $D^3=0,$ so $D=0,$ so $A=0,$ contradicting the fact that the rank is $2.$