Equality case in the Frobenius rank inequality

Question

In many linear algebra books, the following rank inequalities are found:

Frobenius inequality Let $A$, $B$ and $C$ be three matrices such that the product $ABC$ is defined. Then $$\operatorname{rk}(ABC) + \operatorname{rk}(B) \geq \operatorname{rk}(AB) + \operatorname{rk}(BC).$$

In the special case case $B = I$, the Frobenius inequality reduces to the

Sylvester inequality Let $A$ and $B$ be two matrices such that the product $AB$ is defined. Then $$\operatorname{rk}(A) + \operatorname{rk}(B) - n \leq \operatorname{rk}(AB).$$

Now I wonder about the equality cases in those inequalities. It is common knowledge that

In the Sylvester inequality, equality holds if and only if $$\ker(A) \subseteq \operatorname{Im}(B).$$

But I didn't find anything on the Frobenius inequality. So my question is:

How can the equality case in the Frobenius inequality be characterized?

Answer 1

As it was shown by Tian and Styan, equality in the Frobenius rank inequality holds if and only if there exist matrices $X$ and $Y$ (of appropriate sizes) such that $$ BCX+YAB=B. $$In the special case $A=B=C$ this means, that in the Frobenius inequality $rk(A^3)\ge 2rk(A^2)-rk(A)$ we have equality if and only if there exist $X$ and $Y$ such that $A^2X+YA^2=A$. This holds if $A^k=A$ for some $k>1$.