Equality case in the Frobenius rank inequality 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?

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 $k>1$.