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I am trying to figure out whether the following implication is correct or not:

If $B$ is Hermitian and $A$ is Hermitian positive definite, then $A$ and $B$ commute.

Any ideas?

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If it is a claim, then it shouldn't be a question... –  Arturo Magidin Apr 20 '12 at 20:46
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Have you tried this for any pairs of matrices? Just make up some $2\times 2$ Hermitian matrix and some positive definite $2\times 2$ Hermitian matrix, and see whether they commute. –  Jim Belk Apr 20 '12 at 20:50
    
Try a couple examples with $A$ diagonal and $B$ non-diagonal. –  Martin Argerami Apr 20 '12 at 21:35
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And after you have tried some examples, if you then know what's up, you can post it as an answer. –  Gerry Myerson Apr 20 '12 at 23:53
    
Perhaps you mean "if $AB$ is Hermitian ..." –  Robert Israel Apr 21 '12 at 1:16

1 Answer 1

up vote 4 down vote accepted

By the following example, we can show that $A$ and $B$ don't necessarily commute:

Let $A=\begin{bmatrix} 1 & 0\\ 0 &2 \end{bmatrix} $

Let $ B=\begin{bmatrix} 1 &1+i \\ 1-i& 4 \end{bmatrix}$

$A$ is hermitian positive definite, and $B$ is hermitian. Note that: $$AB=\begin{bmatrix} 1 &1+i \\ 2-2i& 4 \end{bmatrix}\neq BA=\begin{bmatrix} 1 &2+2i \\ 1-i & 4 \end{bmatrix}$$

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