Let $A,B$ be $n\times n$ matrices. If $AB=BA$, then $\operatorname{rank}(A^2)+\operatorname{rank}(B^2)\geq2\operatorname{rank}(AB)$.
Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $\operatorname{rank}(B^2)$ is not tractable.
Any hints will be appreciated!
Best Answer
Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix},\quad B=\begin{pmatrix}0&1&1&0\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{pmatrix}.$$ Sorry for misleading. :(