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Why does $H^2U=UH^2$ imply $H$ and $U$ commutes, where $H$ is a Hermitian matrix and $U$ is a unitary matrix?

This comes from the book 'Theory of Matrices' on p277 http://www.maths.ed.ac.uk/~aar/papers/gantmacher1.pdf

Now I know the implication is false. How to prove the following:

If a matrix $A$ is normal, i.e $AA^*=A^*A$, then the polar and unitary factor of the polar decomposition of $A$, $A=UH$, commute.

PS:Actually the book was right. I forgot to mention H is not only Hermitian, H is also positive semi-definite.

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  • $\begingroup$ What do you know about unitary and Hermitian matrices? $\endgroup$ – Umberto P. Sep 17 '15 at 16:28
  • $\begingroup$ @UmbertoP. Only the definition and the polar decomposition of matrices. $\endgroup$ – DDaren Sep 17 '15 at 16:40
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That cannot be true. Take $$ H=\left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) $$ then $H^2=1$ it commutes with anything, however $$U=\left( \begin{array}{cc} 0 & -i \\ i & 0 \\ \end{array} \right)$$ is unitary and does not commute with $H$

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  • $\begingroup$ Sorry a stupid latex typo. Enjoy the counterexample now. $\endgroup$ – Nikolay Gromov Sep 17 '15 at 16:41
  • $\begingroup$ Now +1 deserve this $\endgroup$ – R.N Sep 17 '15 at 16:42
  • $\begingroup$ Good one, +1 for the counterexample. $\endgroup$ – Samrat Mukhopadhyay Sep 17 '15 at 16:45
  • $\begingroup$ Thank you for your example, please see the new edition of the question. $\endgroup$ – DDaren Sep 17 '15 at 16:48
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The book does not say what you are claiming that it says.

This particular $H$ can be written as a polynomial in its square $H^2$, a fact which is true for positive semi-definite matrices (as the book says on the page that you link).

Thus $HU = g(H^2)U = Ug(H^2) = UH.$ The result is not true for arbitrary Hermitian $H$.

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  • $\begingroup$ Yes it was all my fault. There is another restriction on H to make it unique. $\endgroup$ – DDaren Sep 17 '15 at 16:56
  • $\begingroup$ "it's all my fault" -- that's a little harsh, given that the book doesn't prove the polynomial $g$ exists till the next section :-) $\endgroup$ – hunter Sep 17 '15 at 17:04
  • $\begingroup$ I forgot the basic definition of polar decomposition, which is kinda unforgivable. $\endgroup$ – DDaren Sep 17 '15 at 17:07
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Counterexample for part 2 of the question. Let $$ A=UH=\left( \begin{array}{cc} -i & 0 \\ 0 & i \\ \end{array} \right)= \left( \begin{array}{cc} 0 & -i \\ i & 0 \\ \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 1 & 0 \\ \end{array} \right) $$ then $A A^\dagger = A^\dagger A = 1$ but $H$ and $U$ still do not commute.

May be the theorem is that $H$ and $U$ can be chosen to commute, as there could be some freedom in what is $H$ and what is $U$.

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  • $\begingroup$ Sorry it is my fault that I forgot to mention $H$ is also positive semi-definite. I apologise. $\endgroup$ – DDaren Sep 17 '15 at 16:58

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