Polar decomposition of an operator and the partial isometry Let $A \in L(H)$, the space of bounded operators on the Hilbert space $H.$ I need to show that there is a partial isometry $U$ (i.e. $U^{*}U$ and $UU^{*}$ are projections), such that: $$A=U|A|$$ is the polar decomposition of $A,$ whereby $|A|=(A^{*}A)^{1/2}.$
Do you have any suggestion or a solution proposal ? 
Many thanks for any comment.

Let $\sqrt{A^*A}$ denote the unique positive square root of $A^*A$. Then
\begin{align}
    \|Ax\|^2 &=\langle A^*Ax,x\rangle \\ 
    & = \langle \sqrt{A^*A}x,\sqrt{A^*A}x\rangle \\ & =\|\sqrt{A^*A}x\|^2
\end{align}
Let $U$ be defined so that $U=0$ on $\mathcal{N}(\sqrt{A^*A})$ and be defined on $\mathcal{N}(\sqrt{A^*A})^{\perp}=\overline{\mathcal{R}(\sqrt{A^*A})}$ so that $U\sqrt{A^*A}=A$. Then $U$ is a partial isometry, and $A=U|A|$ is the desired decomposition, where $|A|=\sqrt{A^*A}$.
