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Assume, we know that $A,B\in M_{n}(\mathbb R)$. Can we say that:

$$\mathrm{rank}(AB-BA)\leq [n/2]$$

Thanks for any hint.

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    No${}{}{}{}{}{}{}{}{}{}$.2012-12-13
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    why?please introduce an example2012-12-13
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    The title is not descriptive of the question, which doesn't mention the rank of $AB$ or of $A$ or of $B$.2012-12-13
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    @aliakbar: Injam hamin karo bokon. Inha hasas hastan be inke agar jawabi midan, uni ke behtare ro ghabul koni. Khosh waghtam.2012-12-29
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    @aliakbar: Zemnan jomalateto khili ba aadab wa taghaza barashun benewis. Neshun bede ta hala chikar kardi. hamintori nanewis ino jawab bedid ya in chi mishe. Bayad dar morede har kaari neshun bedi ke kami kaar ru mas-ale kardi. Mamnunam ke inja hast.2012-12-29
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    @Babak: Is that... Farsi? I'm interested, because my good friend google.translate doesn't know, but I think I know, which is strange.2012-12-29

2 Answers 2

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For any $n\ge2$, consider the following $n\times n$ permutation matrix and the diagonal matrix $$ A=\begin{pmatrix}&&&1\\1\\&\ddots\\&&1&\end{pmatrix}, \ B=\begin{pmatrix}1\\&2\\&&\ddots\\&&&n\end{pmatrix}. $$ Then $$ AB-BA=\begin{pmatrix}&&&n-1\\-1\\&\ddots\\&&-1&\end{pmatrix} $$ has full rank with determinant $n-1$.

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Hint: try almost any $2 \times 2$ example.