Polar decomposition of an operator and the partial isometry

Question

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.

Answer 1

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.

Answer 2

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 $|A|=\sqrt{A^*A}$.