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In the following theorm, I do not know why $K^\perp = 0$ enter image description here

I just accept that $x = 0$ on $K^\perp$. For instance, $x = h\otimes h$ for $h\in H_{\|.\|=1}$ is a compact operator. Extend $\{h\}$ to a basis $\{e_i\}\cup \{h\}$ for $H$, we have $K^\perp = \bar{span} \{e_i\}$, and $x_{|_{K^\perp}} = 0$ while $K^\perp \neq 0$. Where is my mistake in this example and why $K^{\perp}=0$?

Please regard me. Thanks in advance.

niki
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1 Answers1

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Here is a rephrasing of the argument, with some of the details suppressed and others expanded upon. Maybe it's clear:

Split $H$ between the set "generated" by the eigenvectors of $u$, call it $K$, and the orthogonal complement of that. Since you've used up all the eigenvectors of $u$, the restriction $u_{K^\perp}$ has no eigenvectors. Then by normality, the norm of $u_{K^\perp}$ is equal to its spectral radius, which is $0$ because there are no eigenvalues. (This step is the whole point of the proof, so it bothers me a bit that it is so terse in the argument above.) So $u_{K^\perp}$ itself is zero. Hence anything in $K^\perp$ is either an eigenvector (with eigenvalue zero) or the zero vector; but all the eigenvectors are in $K$, so $K^\perp=\{ 0 \}$.

Ian
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  • I think you objection is a good one. The restriction $u_{K^{\perp}}$ of $u$ to $K^{\perp}$ is compact and normal. It makes more sense to argue that if $u_{K^{\perp}}$ is non-zero then $u_{K^{\perp}}$ has a non-zero eigenvalue. I don't see what would be wrong with that argument instead. Also, what's up with Zorn's lemma? – Disintegrating By Parts Oct 20 '15 at 16:58
  • @TrialAndError There is a pretty standard Zorn's lemma argument for showing that appropriate operators on Hilbert spaces have a maximal orthonormal set of eigenvectors. You only need it to deal with a situation where your operator might have uncountably many linearly independent eigenvectors. – Ian Oct 20 '15 at 18:11
  • Can someone please help me to understand where the compactness property of the operator has been used ? – krishan acton Feb 28 '22 at 18:25
  • @krishanacton For a compact operators the spectrum is ${0} \bigcup {\lambda_n}$ where ${\lambda_n}$ is the set of eigenvalues, since $u_{K^{\perp}}$ does not have eigenvalues then norm must be $0$. – user715747 Apr 26 '23 at 06:47