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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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    @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.