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I need to prove that an operator on a finite dimensional vector space is normal iff given a polar decomposition $P=UH$ with $U$ unitary and $H$ positive semidefinite, $HU=UH$.

If $HU=UH$ a simple calculation shows $T^\ast T=TT^\ast=H^2$, so $T$ is normal. However, I'm having a hard time with the converse. Starting with $TT^\ast=T^\ast T$ I am only able to deduce $UH^2=H^2U$, instead of $UH=HU$. What to do?

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There is a polynomial $p$ such that $p(H^2) = H$. Then $UH^2=H^2U$ implies $UH=Up(H^2)=p(H^2)U=HU$.

The polynomial $p$ should satisfy $p(x)=\sqrt{x}$ for each eigenvalue $x$ of $H^2$. It is then a consequence of diagonalizability of $H$ and nonnegativity of the eigenvalues of $H$ that $p(H^2)=H$. Questions linked in the second comment on the question show more details.

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  • $\begingroup$ Why is there such a polynomial? What's going here? $\endgroup$
    – linalg
    Jul 6, 2016 at 22:10
  • $\begingroup$ @linalg: $H$ and $H^2$ are positive semidefinite. Each eigenvalue of $H^2$ is of the form $\lambda^2$ for an eigenvalue $\lambda$ of $H$, and $\lambda\geq 0$. $H$ and $H^2$ are diagonalizable with the same set of eigenvectors. Polynomial interpolation allows you to specify the values to get square roots, and then noticing how $p(H^2)$ acts on eigenvectors (forming a basis) shows that $p(H^2)=H$. $\endgroup$
    – Jonas Meyer
    Jul 6, 2016 at 22:12
  • $\begingroup$ This is interesting but really not enough details for me to make everything out :\ $\endgroup$
    – linalg
    Jul 6, 2016 at 22:15
  • $\begingroup$ I don't understand still why such a polynomial exists.. $\endgroup$
    – linalg
    Jul 6, 2016 at 22:16
  • $\begingroup$ That is a rather broad statement, and I do not know if it is just the polynomial interpolation part you are unclear on, or why the interpolating polynomial does what is needed. If the former, you can Google "polynomial interpolation." For the latter, it is the spectral theorem. $\endgroup$
    – Jonas Meyer
    Jul 6, 2016 at 22:18

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