polar decomposition proof

Question

Let $H$ be a hilbert space and $T$ a bounded linear operator on $H$. I'm trying to prove that there is a partial isometry $V$ on the closure of $Im(|T|)$ such that $T=V|T|$ and $|T|=V^*T$, where $|T|={\sqrt{T^*T}}$. I've proved the first equality. For the second, I took the adjoint of both sides of first and post multiplied by $T$ to get $|T|^2=|T|V^*T$ so that rearranging we get $|T|(|T|-V^*T)=0$. All this tells me is that $|T|$ is zero on $Im(|T|-V^*T)$. How do I deduce that $|T|=V^*T$?

Thanks

Answer 1

We have $T=V|T|$. Left multiply both sides by $V^{*}$ to get $V^{*}T=V^{*}V|T|$. Since $V$ is a partial isometry on the closure of the range of $|T|$, $V^{*}V=I$ on Im $|T|$ whence $V^{*}T=|T|$.

Answer 2

The notation $|T|$ means the unique positive square root of $T^{\star}T$. Notice that $$ \||T|x\|^{2}=(|T|x,|T|x)=(|T|^{2}x,x)=(T^{\star}Tx,x)=(Tx,Tx)=\|Tx\|^{2}. $$ So $\mathcal{N}(T)=\mathcal{N}(|T|)$. Let $Y=\mathcal{R}(|T|)^{c}$. Then $|T| : Y\rightarrow Y$ is invertible because $\mathcal{N}(|T|)\cap Y=\mathcal{N}(|T|)\cap\mathcal{N}(|T|)^{\perp}=\{0\}$. Let $V=T|T|^{-1}$ on $\mathcal{R}(|T|)$ and let $V=0$ on $Y^{\perp}$. Then $$ \|Vy\|=\|T|T|^{-1}y\|=\||T||T|^{-1}y\|=\|y\|,\;\;\; y \in \mathcal{R}(|T|). $$ So $V$ has a unique continuous extension to all of $X$, and that extension is a partial isometry: $V$ is isometric on $Y$ and $0$ on $Y^{\perp}$. And $V^{\star}V=P$ is the orthogonal projection onto $Y$. By definition of $V$, one has $T=V|T|$. Because $P|T|=|T|$, $$ V^{\star}T = V^{\star}V|T|=P|T|=|T|. $$

Answer 3

I misinterpreted your question. I thought you wanted to see the standard construction of the $V$ where $V^{\star}V=P$ is the orthogonal projection onto $\mathcal{R}(|T|)^{c}$. This $V$ is unique in all cases, even where $\mathcal{N}(T)\ne \{0\}$, whereas a general partial isometry $V$ is not uniquely determined if $T$ has a non-trivial null space.

$V$ is uniquely determined on $\mathcal{R}(|T|)^{c}$ by the requirement that $T=V|T|$. That' I've shown you because $V=T|T|^{-1}$ is isometric on $\mathcal{R}(|T|)$ and extends uniquely by continuity to an isometry on the closure $\mathcal{R}(|T|)^{c}$. However, you can set $V$ to anything on $\mathcal{N}(|T|)=\mathcal{R}(|T|)^{\perp}$ and still have $T=V|T|$, even though $V$ may no longer be a partial isometry on the full space; but an arbitrary extension of $V$ isn't good for the adjoint equation, though an isometric extension one is.

If you extend $V$ to a partial isometry on the entire space, then $V^{\star}V=P$ is an orthogonal projection onto some subspace which will at least include $\mathcal{R}(|T|)^{c}$. Any such extension $V$ will satisfy $V^{\star}T=V^{\star}V|T|=P|T|=|T|$ because $P=I$ on $\mathcal{R}(|T|)^{c}$. The $V$ I gave you is minimal in the sense that $V=0$ on $\mathcal{R}(|T|)^{\perp}=\mathcal{N}(|T|)$--it is uniquely determined by $T=V|T|$ and $V=0$ on $\mathcal{N}(|T|)$.

The ambiguity you determined concerning the null space of $|T|$ is the issue.