# In Uhlmann's theorem, should the polar decomposition be written as $A=|A|V$ or $A=V|A|$?

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 $$A^\dagger A = |A|V^\dagger V|A| = |A||A| = A^\dagger A$$.

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

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 $$P=\sqrt{AA^\dagger}$$.