# Rank product of Positive Semidefinite Matrix and a Full Rank Matrix

$$\newcommand{\rank}{\operatorname{rank}}$$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, $$AB\neq {\bf 0}$$.

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.

• Thank you such much for your answer! Oct 13, 2020 at 17:16
• What happens if $AB$ is not a null matrix? Oct 13, 2020 at 17:37
• You can add a column to $B$ that is linearly independent of $b$. Then $B$ will have rank 2, but $AB$ will have rank $1$ (the first column will be zero as before). The idea is that you make $A$ rank-deficient in one of the dimensions of $B$, eliminating that dimension in the product. Oct 13, 2020 at 18:00
• Thank you for you answer! Oct 13, 2020 at 18:02

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 $$\operatorname{rk}(AB)\ge\max\{0, r+\operatorname{rk}A-n\}$$.

• Thank you such much for your answer! Oct 13, 2020 at 17:16
• What happens if $AB$ is not a null matrix? Oct 13, 2020 at 17:37
• @JuanPabloSotoQuirós Nothing, in the sense that, for all matrices $A\in\Bbb F^{n\times h}$ and natural numbers $m$, $s$ and $t$ such that $0\le s\le\min\{m, h\}$ and $\max\{0,s+\operatorname{rk}A-h\}\le t\le\min\{\operatorname{rk}A,s\}$, there are matrices $B\in\Bbb F^{h\times m}$ such that $\operatorname{rk}(AB)=t$ and $\operatorname{rk}B=s$.
– user239203
Oct 13, 2020 at 20:36