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.

You will need to work out the details, but basically you want to define an operator $U$ on (closure of) the range of $|A|$ by 
$$
U(|A|x)=Ax 
$$
and zero on $(AH)^\perp$. 
The only non-trivial part, I think, is to show that this is well defined. But this works, because  if $Ax=Ay$ then $A^*Ax=A^*Ay$ and now you can extend this equality to polynomials, take limits, and get $|A|x=|A|y$. 
Next show that $U$ is an isometry on its domain, so it is a partial isometry, and in particular bounded. 
