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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?

linalg
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  • Related: http://math.stackexchange.com/questions/1439766/why-h2u-uh2-implies-h-and-u-commutes?rq=1 – Jonas Meyer Jul 06 '16 at 22:10
  • See also: http://math.stackexchange.com/questions/114171/hermitian-positive-semi-definite-square-root, http://math.stackexchange.com/questions/134186/does-certain-matrix-commutes-with-square-root-of-another-one, – Jonas Meyer Jul 06 '16 at 22:37
  • @JonasMeyer thanks! – linalg Jul 06 '16 at 22:42

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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.

Jonas Meyer
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  • Why is there such a polynomial? What's going here? – linalg Jul 06 '16 at 22:10
  • @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 $p(H^2)=H$. – Jonas Meyer Jul 06 '16 at 22:12
  • This is interesting but really not enough details for me to make everything out :\ – linalg Jul 06 '16 at 22:15
  • I don't understand still why such a polynomial exists.. – linalg Jul 06 '16 at 22:16
  • 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. – Jonas Meyer Jul 06 '16 at 22:18
  • My question is "why does there exist a polynomial $p$ such that $p(H^2)=H$? The only spectral theorems I know just give sufficient conditions for diagonalization, and I've never seen anything like what you wrote :( – linalg Jul 06 '16 at 22:22
  • linalg: There are several ingredients working together here. It is OK if you haven't seen this combination of them before. Diagonalize H, with eigenvalues $\lambda_1,\ldots,\lambda_n$ (possibly with repetition). Then $H^2$ is the operator with the same eigenvectors and respective eigenvalues $\lambda_1^2,\ldots,\lambda_n^2$. If $p$ is a polynomial such that $p(\lambda_k^2)=\lambda_k$ for each $k$, which is possible because each $\lambda_k$ is nonnegative (google interpolating polynomial if you haven't yet), then $p(H^2)$ will have same eigenvectors with same eigenvalues as $H$, hence $=H$. – Jonas Meyer Jul 06 '16 at 22:27
  • @linalg: I just searched the site and found this answer with more explanation: http://math.stackexchange.com/a/134190/1424 – Jonas Meyer Jul 06 '16 at 22:34
  • Thanks! Maybe a silly question, but why does having the same eigenvalues and eigenvectors imply equal operators? – linalg Jul 06 '16 at 22:41
  • Does this depend on being diagonalizable, since then we know the action on a basis, which determines an operator? – linalg Jul 06 '16 at 22:42
  • @linalg: Yes, that is it. – Jonas Meyer Jul 06 '16 at 22:42