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.

  • $\begingroup$ 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. $\endgroup$ – 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.


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