In the proof of Uhlmann's theorem, the book writes the polar decomposition: $A = |A|V$, with $|A| = \sqrt{A^\dagger A}$.

Shouldn't it be $V|A|$ instead? The former case is $A^\dagger A = V^\dagger|A||A|V$ while the latter case is $A^\dagger A = |A|V^\dagger V|A| = |A||A| = A^\dagger A$.

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    $\begingroup$ mathjax is supported out of the box, you don't need to add it manually $\endgroup$ – glS Nov 30 '20 at 11:18

Ok I confused myself in the first version of this answer.

With the definition $|A|:=\sqrt{A^\dagger A}$, it should be $A=U|A|$. If you define instead $|A|:=\sqrt{A A^\dagger}$, then it is $A=|A|U$. Both definitions are common.

There is a left and a right polar decomposition $A=PU=UP'$ which are related to the SVD $A=V\Sigma W^\dagger$ by $$ P=V\Sigma V^\dagger, \quad U = VW^\dagger, \quad P' = W\Sigma W^\dagger. $$ Note that $\Sigma$ is the diagonal matrix with the singular values of $A$, i.e. the eigenvalues of $|A|=\sqrt{A^\dagger A}$ (or $\sqrt{A A^\dagger}$, they are the same), on its diagonal. Clearly, we have $$ A^\dagger A = P'U^\dagger U P' = W\Sigma^2 W, \quad AA^\dagger = V\Sigma^2 V^\dagger $$ hence $P'=\sqrt{A^\dagger A}=|A|$ and $P=\sqrt{AA^\dagger}$.


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