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In general when one proves that the product of semisimple (i.e. diagonalizable) matrices is semisimple one assumes they commute and are thus simultaneously diagonalizable, and then the result follows. I was wondering if anyone knew of an example of non-commuting semisimple matrices whose product is not semisimple.

flag	asked Mar 8 2011 at 7:30 HNuer 915●2●7

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The question only involves some basic linear algebra, so the tag algebraic-groups is not needed here. Also, "semisimple" refers to a matrix over an arbitrary field (perhaps not containing its eigenvalues) which is diagonalizable over some field extension. The term "diagonalizable" requires the qualification "over some extension field" to avoid confusion unless you refer to a matrix which has eigenvalues in the given field. – Jim Humphreys Mar 8 2011 at 15:26

Bruce Westbury · Accepted Answer · 2011-03-08 08:34:44Z UTC

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How about $\begin{pmatrix} 5 & 3 \\ 8 & 5 \end{pmatrix}$ and $\begin{pmatrix} 2 & -3 \\ -3 & 5 \end{pmatrix}$?

answered Mar 8 2011 at 8:34

Bruce Westbury
6,635●2●10●25

great! thanks a lot. this is a good example, since i forgot to mention that i wanted them to be invertible as well. – HNuer Mar 8 2011 at 8:46

Amritanshu Prasad · Answer 2 · 2011-03-08 08:39:14Z UTC

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$\begin{pmatrix}1&0\\0&0\end{pmatrix}\begin{pmatrix}0&1\\1&0\end{pmatrix}=\begin{pmatrix}0 & 1\\0&0\end{pmatrix}$

answered Mar 8 2011 at 8:39

Amritanshu Prasad
1,873●6●15

The product of non-commuting semisimple matrices need not be semisimple

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