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For matrices $A, B$, does the following rank (rk) inequality hold \begin{equation} {\rm rk} (AB + BA) \le {\rm rk}(AB) + {\rm rk}(BA) \le {\rm min}({\rm rk} A, {\rm rk} B) + {\rm min}({\rm rk} B, {\rm rk} A) = 2 ~{\rm min}({\rm rk} A, {\rm rk} B) ? \end{equation}

Rob
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    Yes, you just first apply the inequality $\text{rk}(A + B) \le \text{rk}(A) + \text{rk}(B)$ and then apply the inequality in the title. – Qiaochu Yuan Oct 08 '22 at 23:46

2 Answers2

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For Rob's extra question.
Try these: $$ A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \qquad B = \begin{bmatrix} 0 & 0 \\ 1 & 0 \end{bmatrix} $$ both of rank $1$. Then $$ AB = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \qquad BA = \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $$ and $AB+BA$ has rank $2$.

GEdgar
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Yes, because the following two inequalities hold:

(1) The one in the title,

(2) ${\rm rk}(X+Y) \le {\rm rk}(X) + {\rm rk}(Y)$.

Apply the two inequalities and you can prove the one in your question.

Lapin
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  • Thanks, @Lapin. How about the maximum value of $rk (AB+BA)$ if both $rk A = rk B =1$? I mean I checked numerically with several examples and I dont find it greater than 1. – Rob Oct 09 '22 at 00:29
  • @Rob Can you get $2\times 2$ matrices $A$ and $B$ of rank $1$ so that $AB$ has a $1$ in the upper left but $BA$ has a $1$ in the lower right? – GEdgar Oct 09 '22 at 00:37
  • @GEdgar, doesn't seem so! You have any examples where $rk (AB+BA) > 1$ for $rk A = rk B = 1$? – Rob Oct 09 '22 at 00:47