Rank product of Positive Semidefinite Matrix and a Full Rank Matrix

Question

$\newcommand{\rank}{\operatorname{rank}}$Given two matrices $A\in\mathbb{R}^{n\times n}$ and $B\in\mathbb{R}^{n\times r}$, I known that $$\rank(AB)\leq\min\{\rank(A),\rank(B)\}.$$

If $n\geq r$, $A$ is a positive semi-definite matrix and $B$ is full rank, i.e., $\rank(B)=r$, can I said that $$\rank(AB)=\min\{\rank(A),\rank(B)\}\text{?}$$ Any information will be helpful.

Edit: Also, $AB\neq {\bf 0}$.

Answer 1

Let $b\in\mathbb{R}^n$ be a vector satisfying $b^\top b = 1$. Let

$$A = I - bb^\top, \qquad B = b.$$

Then, $B\in\mathbb{R}^{n\times 1}$ has full rank (equal to $1$), $A$ is positive semidefinite with rank $n-1$ and all eigenvalues equal to zero or $1$, and

$$AB = b - b = 0,$$

which has zero rank.

Answer 2

No, consider for instance $A=\begin{pmatrix}1&0\\ 0&0\end{pmatrix}$, $B=\begin{pmatrix}0\\1\end{pmatrix}$. However, for two matrices $A\in \Bbb F^{n\times h}$, $B\in\Bbb F^{h\times m}$ you always have the inequality $$\operatorname{rk}(AB)\ge \operatorname{rk}B-\dim\ker A=\operatorname{rk}B+\operatorname{rk}A-h$$ This translates, in your instance, to $\operatorname{rk}(AB)\ge\max\{0, r+\operatorname{rk}A-n\}$.