Prove that the polar decomposition of normal matrices, $A=SU$, is such that $SU=US$

Question

Assume $A$ is a normal matrix. Suppose $A=SU$ is a polar decomposition of $A$. Prove that $SU=US$.

I have no idea to prove this.

$A$ is normal then $AA^*=A^*A$. And then we have $$ SS^*=U^*S^*SU. $$ But I don't know how to continue.

Answer 1

Let $\,A=U|A|$, then $\,A^*=|A|U^*$. By normality one obtains $$U|A|^2U^* = AA^* = A^*A = |A|^2,$$ an equality of positive-semidefinite matrices.

"Positive square-rooting" yields $\,U|A|U^* = |A|\;\Longleftrightarrow\; U|A| = |A|U$.

Answer 2

Hint: Note that, since $A = SU$ is a polar decomposition, $S$ is (Hermitian and) positive semidefinite.

So, as you noted, we have $$ SS^* = U^*S^*SU \implies\\ S^2 = U^* S^2U $$ From here, note that each side is positive semidefinite and that the positive semidefinite square root of such a matrix is uniquely defined. As such, we can take the square root of both sides.

Answer 3

Let $\newcommand{\bs}[1]{\boldsymbol{#1}} A$ be a normal matrix. It is therefore unitarily diagonalisable, i.e. writable as $$A = \sum_k c_k P_{\bs u_k}$$ for a set of orthogonal projectors $P_{\bs u_k}\equiv \bs u_k \bs u_k^*$, where $\langle \bs u_j,\bs u_k\rangle=\delta_{jk}$ (and thus $\operatorname{Tr}(P_{\bs u_j} P_{\bs u_k})=\delta_{jk}$) and complex coefficients $c_k\in\mathbb C$.

The corresponding SVD (or more generally one possible choice of SVD) reads $A=UD V^\dagger$ with $$ U = \sum_k e^{i\varphi_k} \bs u_k \bs e_k^*,\qquad D=\sum_k |c_k| \bs e_k\bs e_k^*, \qquad V=\sum_k \bs u_k \bs e_k^*, $$ where $c_k\equiv |c_k|e^{i\varphi_k}$, and $\bs e_k$ are the basis vectors for some reference basis used to represent the matrices.

The polar decomposition reads $A=|A| U_A$ with $$|A|\equiv UDU^\dagger= \sum_k |c_k| \bs u_k \bs u_k^*, \qquad U_A \equiv UV^\dagger = \sum_k e^{i\varphi_k} \bs u_k \bs u_k^*.$$ From this, it's clear that $|A|$ and $U_A$ commute, being them diagonal in the same basis.

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It's worth seeing where exactly the above argument breaks in the case of non-normal matrices. For a generic matrix $A$ the SVD reads $$A = UDV^\dagger = \sum_k p_k \bs u_k \bs v_k^*, \\ U\equiv \sum_k \bs u_k\bs e_k^*, \qquad V \equiv \sum_k \bs v_k\bs e_k^*, \qquad D \equiv \sum_k p_k \bs e_k \bs e_k^*,$$ for some positive coefficients $p_k>0$ and orthonormal sets $\{\bs u_k\}_k$ and $\{\bs v_k\}_k$. Then $A=|A| U_A$ with $$|A| \equiv UD U^\dagger = \sum_k p_k \bs u_k \bs u_k^*, \qquad U_A \equiv UV^\dagger = \sum_k \bs u_k \bs v_k^*. $$ Now we have $$U_A |A| = (UV^\dagger U)DU^\dagger = \sum_{jk} p_k \langle \bs v_j,\bs u_k\rangle \bs u_j \bs u_k^*,$$ which in general differs from $|A| U_A=A$.