# Normal matrix on f.d space implies unitary and positive semidefinite matrices commute

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?

• Jul 6, 2016 at 22:10
• Jul 6, 2016 at 22:37
• @JonasMeyer thanks! Jul 6, 2016 at 22:42

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.
• @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 $p(H^2)=H$. Jul 6, 2016 at 22:12