Does equality hold in Frobenius inequality if $XZ=I$

Question

The Frobenius inequality states that: $$\operatorname{rank}(XY)+\operatorname{rank}(YZ) \le \operatorname{rank}(Y)+\operatorname{rank}(XYZ)$$ My question: Does the equality hold in Frobenius inequality if $XZ=I$

Answer 1

If $X$ and $Z$ are nonsingular matrices, the answer is "yes", because $$ \operatorname{rank}(XY)=\operatorname{rank}(YZ)=\operatorname{rank}(XYZ)=\operatorname{rank}(Y). $$ If $X$ and $Z$ are not square matrices and $Z$ is merely a right inverse of $X$, the answer is "no". E.g. try $$ X=\pmatrix{1&0},\ Y=\pmatrix{0&0\\ 0&1},\ Z=X^T. $$