# Polar decomposition of normal matrix

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.

• This post deals with "$\!A$ is a normal operator" $\Rightarrow$ "The factors in the polar decomp. of $A$ commute." Note that normality is also a necessary condition, so that both conditions are equivalent in fact. This is subject of a recent post which refers to an exercise in J. Conway's book on Functional Analysis. – Hanno Mar 19 at 11:06

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 $$\,U|A|U^* = |A|\;\Longleftrightarrow\; U|A| = |A|U$$.
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.