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I am trying to prove if H is a separable Hilbert space then the operator T in B(H) has the form U|T| (where U is a unitary operator and $|T|=(T^* T)^1/2$ ) iff index T=0. I was thinking to start with the polar decomposition of T i.e $T=U|T|$ but how should i proceed to index T=0.

Nina
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For more see: Construction, Example, Overview

Hilbert Decomposition: $$X=\overline{\mathcal{R}|T|}\oplus\overline{\mathcal{R}|T|}^\perp\quad Y=\overline{\mathcal{R}T}\oplus\overline{\mathcal{R}T}^\perp$$

Square Root Lemma: $$T\in B(X,Y):\quad\left\|Tx\right\|=\left\||T|x\right\|$$ Unitary Operator: $$U:\overline{\mathcal{R}|T|}\leftrightarrow\overline{\mathcal{R}T}:\quad U|T|\varphi:=T\varphi$$

Range Relation: $$\overline{\mathcal{R}|T|}^\perp=\mathcal{R}|T|^\perp=\mathcal{N}|T|=\mathcal{N}T$$

Dimension Criterion: $$T=U|T|\iff\dim\mathcal{N}T=\dim\mathcal{R}T^\perp$$

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C-star-W-star
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  • How can i combine all these to het a proof?? – Nina Sep 03 '16 at 12:30
  • @Nina: Basically you need to construct a unitary $U$ between $X$ and $Y$ such that $U|T|x=Tx$ for all $x\in X$. For this you start with an isometry between the ranges $\mathcal{R}|T|$ and $\mathcal{R}T$. This works because of the square root lemma! That you can extend uniformly to obtain a unitary between their closures. So far so good! Now you proceed with the parts orthogonal to these ranges: They are unitarily equivalent iff their dimension agree! A few minor relations between ranges and kernels lets you rephrase this in a more comfortable fashion. Does this help? Let me know! :) – C-star-W-star Sep 03 '16 at 18:44
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    @C-Star-Puppy, Can you explain both inclusions in Dimension Criterion. – Sumanta Feb 19 '19 at 17:02
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    @Mathlover, sure. $U$ is already a unitary between $\overline{\mathcal{R}|T|}$ and $\overline{\mathcal{R}T}$. In order to extend this to a unitary between the full spaces $X=\overline{\mathcal{R}|T|}\oplus\mathcal{R}|T|^\perp$ and $Y=\overline{\mathcal{R}T}\oplus\mathcal{R}T^\perp$ we require a unitary between the missing parts $\mathcal{R}|T|^\perp$ and $\mathcal{R}T^\perp$. But this is the case if and only if these are isomorphic which is the case if and only if they have the same dimension. Rephrased by the the "Range Relation" this gives the "Dimension Criterion". – C-star-W-star Feb 21 '19 at 11:08
  • @Mathlover: In case you have more questions feel free to ask. – C-star-W-star Feb 21 '19 at 11:35