How do you prove that every matrix $A\in\mathbb{C}^{n\times n}$ has a polar decomposition, i.e. can be written as $A=UP$, $U$ being unitary ($U^\ast U=UU^\ast=I$, with $U^\ast$ indicating the conjugate transpose of $U$) and $P$ positive semi-definite Hermitian? I have seen this link on Wikipedia, but I am not convinced that $U$ is unitary, since if $U=AP^{-1}$, then:
$$UU^\ast=AP^{-1}(AP^{-1})^\ast=AP^{-1}(P^{-1})^\ast A^\ast,$$
which even swapping $^{-1}$ with $^\ast$ on $(P^{-1})^t$ doesn't seem to lead to the identity matrix $I$. Any suggestions on that?
P.S. Why does [this link on Wikipedia](en.wikipedia.org/wiki/Polar_decomposition) not produce "this link on Wikipedia" with hyperlink to the given URL? Shouldn't it? Because you must put http:// for it to work, or the link won't be parsed.
P.P.S. I actually only need the statement for real-entry matrixes, i.e. $A\in\mathbb{R}^{n\times n}$, where $^\ast\leadsto^t$ and unitary$\leadsto$orthogonal, and Hermitian$\leadsto$symmetric.
Best Answer
I give here the real version of your equality. Notice that the matrix $A^tA$ is symmetric positive semi-definite then it's diagonalizable and its eigenvalues are non negative: $$A^tA=QDQ^{-1}$$ where $D=\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$ and we can define the square root of $A^tA$ denoted by $\sqrt{A^t A}$ by $$\sqrt {A^tA}=Q\sqrt D Q^{-1}$$ where $$D=\operatorname{diag}(\sqrt\lambda_1,\ldots,\sqrt\lambda_n)$$ so let $P=\sqrt {A^tA}$ and $O=AP^{-1}$ then it's clear that $P$ is symmetric positive semi-definite and we have $$O^tO=(P^{-1})^tA^tAP^{-1}=(P^{-1})^tP^2P^{-1}=I$$ so $O$ is orthogonal.