Compact operators

**Dinkydoe** · December 22nd 2009, 10:20 AM

I'm struggeling with the following questions:

Prove or disprove the following statements:
(a) If $\mathcal{H}$ is a complex Hilbert-space and $T\in B(\mathcal{H})$ be a compact operator such that $\sigma(T) = \left\{0\right\}$ (the spectrum) then $T= 0$
(b) If $\mathcal{H}$ is a complex Hilbert-space and $T\in B(\mathcal{H})$ such that $T^2$ is compact, then the set $\left\{\lambda\in\sigma(T):|\lambda|>\delta \right\}$ is finite for each $\delta>0$ .

I suspect (a) is false: Taking a Hilbert-space of finite dimension, and a nilpotent matrix should be a counterexample.
I suspect (b) is false. If we find a compact operator $T$ such that $\sigma(T)$ is (not countable) infinite, this should be a counter example for (b). However I'm not even sure such T exists.

Any help would be appreciated!

**Opalg** · December 23rd 2009, 08:03 AM

Originally Posted by Dinkydoe

I'm struggeling with the following questions:

Prove or disprove the following statements:
(a) If $\mathcal{H}$ is a complex Hilbert-space and $T\in B(\mathcal{H})$ be a compact operator such that $\sigma(T) = \left\{0\right\}$ (the spectrum) then $T= 0$
(b) If $\mathcal{H}$ is a complex Hilbert-space and $T\in B(\mathcal{H})$ such that $T^2$ is compact, then the set $\left\{\lambda\in\sigma(T):|\lambda|>\delta \right\}$ is finite for each $\delta>0$ .

I suspect (a) is false: Taking a Hilbert space of finite dimension, and a nilpotent matrix should be a counterexample. (Correct.)
I suspect (b) is false. If we find a compact operator $T$ such that $\sigma(T)$ is (not countable) infinite, this should be a counter example for (b). However I'm not even sure such T exists.

You can get a counterexample to (a) on an infinite-dimensional space by taking T to act nilpotently on a finite-dimensional subspace, and to be zero on the orthogonal complement.

But (b) is true. Hint: $T^2-\lambda^2 I = (T+\lambda I)(T-\lambda I)$ . If $T^2-\lambda^2 I$ is invertible then so is $T-\lambda I$ .

**Dinkydoe** · December 23rd 2009, 09:15 AM

So if I'm right, and please correct me where I'm wrong:

There's a Theorem in my book that more or less states: For compact operators we have $\sigma_p(T)=\sigma(T)$ (Pointspectrum = Spectrum). And another Theorem states that (*)for any t > 0 the set of all distinct eigenvalues $\lambda$ of $T$ with $\lambda \geq t$ is finite.

Thus given your hint: if $\lambda = a^2$ is such that $T^2-\lambda = (T-a)(T+a)$ is invertible iff $T\pm a$ are invertible

Thus $\lambda\in \rho(T^2) \Leftrightarrow \pm a \in \rho(T)$ . And since $T^2$ is compact we have (*), thus since $\sigma(T)= \mathbb{C}\setminus \rho(T)$ should follow that for any $\delta >0$ , we have $\left\{a\in\sigma(T):|a|\geq \delta \right\}$ is finite.

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