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Let $A,B$ be $n\times n$ matrices. If $AB=BA$, then $\operatorname{rank}(A^2)+\operatorname{rank}(B^2)\geq2\operatorname{rank}(AB)$.

Is this rank inequality correct? No counterexample seems to exist. Here's what I've done. When $A$ is a Jordan block this is easily proved. I tried to bring $A$ into Jordan form for $n=2,3,4$ and found no counterexample. So currently I think it is true. By Fitting's lemma it suffices to consider the case in which $A$ is nilpotent. We can bring $A$ into Jordan form. Then the blocks in $B$ are upper triangular. However, even in this case $\operatorname{rank}(B^2)$ is not tractable.

Any hints will be appreciated!

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    $\begingroup$ I don't know but it's funny, that this looks very much alike $$ {\rm rank}(A^2 + B^2 - 2AB) = {\rm rank}((A-B)^2) \geq 0 $$ if $AB=BA$ ;) . $\endgroup$ – Diger Nov 26 '18 at 13:57
  • $\begingroup$ What about $${\rm rank}(A^2 + B^2 - 2AB) \leq {\rm rank}(A^2) + {\rm rank}(B^2) + {\rm rank}(-2AB) \, ? $$ $\endgroup$ – Diger Nov 26 '18 at 14:36
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    $\begingroup$ @Diger You got me! It was inspired by the famous inequality $a^2+b^2\geq2ab$. :) $\endgroup$ – Colescu Nov 26 '18 at 14:43
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Well, turns out I made a mistake when trying to construct counterexamples... There is indeed a counterexample of order $4$: $$A=\begin{pmatrix}0&1&0&0\\0&0&0&0\\0&0&0&1\\0&0&0&0\end{pmatrix},\quad B=\begin{pmatrix}0&1&1&0\\0&0&0&1\\0&0&0&1\\0&0&0&0\end{pmatrix}.$$ Sorry for misleading. :(

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  • $\begingroup$ Thanks anyway.I've been doing this a whole day. $\endgroup$ – Oolong milk tea Nov 27 '18 at 12:57
  • $\begingroup$ Now as you have presented counterexample it seems very logical: rank$(A^2)=0$, rank$(B^2)=1$ and if only $AB=BA \ne 0$ then counterexample is ready.. $\endgroup$ – Widawensen Nov 27 '18 at 13:05
  • $\begingroup$ We see that the counterexample is of the form $AB=\begin{bmatrix}N & I \\ 0 & N \end{bmatrix}\begin{bmatrix}N & 0 \\ 0 & N \end{bmatrix}=\begin{bmatrix}0 & N \\ 0 & 0 \end{bmatrix}=BA$ with $\begin{bmatrix}N & I \\ 0 & N \end{bmatrix}^2=\begin{bmatrix}0 & 2N \\ 0 & 0 \end{bmatrix}$ and $\begin{bmatrix}N & 0 \\ 0 & N \end{bmatrix}^2=\begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix}$. Maybe it's worth to memorize for future questions :) .. Thank you for the inspiring question. $\endgroup$ – Widawensen Nov 27 '18 at 14:12

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