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

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2 Answers 2

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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.

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  • $\begingroup$ Thank you such much for your answer! $\endgroup$ Oct 13, 2020 at 17:16
  • $\begingroup$ What happens if $AB$ is not a null matrix? $\endgroup$ Oct 13, 2020 at 17:37
  • $\begingroup$ 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. $\endgroup$ Oct 13, 2020 at 18:00
  • $\begingroup$ Thank you for you answer! $\endgroup$ Oct 13, 2020 at 18:02
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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 $\operatorname{rk}(AB)\ge\max\{0, r+\operatorname{rk}A-n\}$.

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  • $\begingroup$ Thank you such much for your answer! $\endgroup$ Oct 13, 2020 at 17:16
  • $\begingroup$ What happens if $AB$ is not a null matrix? $\endgroup$ Oct 13, 2020 at 17:37
  • $\begingroup$ @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 $\operatorname{rk}B=s$. $\endgroup$
    – user239203
    Oct 13, 2020 at 20:36

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