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I am working on a homework problem (Analysis now E3.3.7) and I have no idea on how to solve it. Can anyone give some thoughts? Many thanks.

Assume that Hilbert pace $H$ is separable and prove that an operator $T$ in $B(H)$ has the form $U|T|$ for some unitary operator $U$ with the $|T|=(T^*T)^{\frac{1}{2}}$ iff index$T = 0$.

I know if $T$ is invertible, then $U$ in polar decomposition is unitary. But how to prove $T$ is invertible if index$T = 0$. And I have no idea on the other direction.

Yuanchen Li
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Remaining spaces: $$\overline{\mathcal{R}|A|}^\perp=\mathcal{N}|A|=\mathcal{N}A\quad\mathcal{N}A^*=\overline{\mathcal{R}A}^\perp$$

For equal dimensions: $$\dim\mathcal{N}A=\dim\mathcal{N}A^*\implies UU^*=1$$

For more details: Polar Decomposition

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C-star-W-star
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  • thanks! But could you explain a little more? I still cannot figure it out. – Yuanchen Li Apr 10 '16 at 04:05
  • Sure, so you have already a unitary between the spaces $\overline{\mathcal{R}|A|}$ and $\overline{\mathcal{R}A}$. So the remaining part of the Hilbert spaces is $\overline{\mathcal{R}|A|}^\perp$ resp. $\overline{\mathcal{R}A}^\perp$. Besides remind then that $\mathcal{S}^\perp=\overline{\mathcal{S}}^\perp$, that $\mathcal{R}T^\perp=\mathcal{N}T^$ and especially $|A|=|A|^$. But they are unitarily equivalent in the sense of Hilbert spaces exactly iff their dimensions agree. You can then glue both unitary operators together without worrying about the Polar decomposition relation. – C-star-W-star Apr 10 '16 at 10:47
  • Many thanks. I think I get it. – Yuanchen Li Apr 10 '16 at 18:19