Rank$(AB) <$ Rank$(B)$ when $A,B,AB$ are nilpotent matrices

Question

Let $A,B$ be $n\times n$ matrices, then we have inequality $\operatorname{rank}(B) \geq \operatorname{rank}(AB)$. I was wondering whether it is possible to make this inequality strict if $A,B,AB$ are sufficiently singular, in particular by making them nilpotent.

That is if we also have the properties $A^n = B^n = (AB)^n = 0$, does it follow that $\operatorname{rank}(B) > \operatorname{rank}(AB)$ holds strictly?

Answer 1

Consider $A = \begin{bmatrix}0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0\end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 &0 \\ 0&0&1 \\ 0&0&0 \end{bmatrix}.$

Then we have $AB = \begin{bmatrix}0&0&1 \\ 0&0&0 \\ 0&0&0\end{bmatrix}.$

So you don't have strict inequality.