4
votes
2answers
61 views

Question about a counterexample concerning compact operators

Does anybody know if the following is true, Let $H$ be an infinite dimensional Hilbert-space and $K:H\rightarrow H$ a compact operator. Then if $|\mathrm{spec}(K)|<\infty$ i.e the spectrum is ...
4
votes
1answer
78 views

What makes compact operators special?

I would like to understand why compact operators are considered so special to consider them as an extra class of operators. Over Hilbert spaces these (as far as I know) these are the ones with ...
1
vote
3answers
39 views

Examples of spectrum of compact operators on the sequence space $l_2$

Suppose $T$ is a compact operator on the sequence space $l_2$, and let $\sigma(T)$ be its spectrum. Is it possible to find a $T \ne 0$ such that $\sigma(T) = \{0\}$? Also, is it possible to find $T$ ...
0
votes
1answer
64 views

The eigenvalues of a compact and self-adjoint operator on Hilbert space

Show that if $K$ is a compact self-adjoint operator on Hilbert space then it has either finitely many eigenvalues or a sequence of eigenvalues $\lambda_n\to 0$ as $n\to \infty$.
0
votes
0answers
15 views

characterisation up to unitary equivalence

My book says that the spectral theorem for compact normal operators characterises compact normal operators up to unitary equivalence. It doesn't expand on this so I was wondering what does this mean ...
7
votes
3answers
113 views

Spectrum of a compact operator

If the spectrum of a compact operator is finite, I don't understand why $0$ has to be a member. I have proved that for all $\epsilon > 0$, there is only a finite number of eigenvectors which have ...
2
votes
1answer
94 views

Direct sum of eigenspaces of a compact operator has finite codimension

In an infinite dimensional Hilbert space the orthogonal complement of the (closure) of the direct sum of eigenspaces of a compact normal operator is finite dimensional. Why is this the case? thanks.
2
votes
1answer
118 views

Show for compact operator $K$, if $||Kf|| < ||f|| \forall f$, then $||K|| < 1$.

I wanted to check my reasoning on proving this statement, and see if anyone had suggestions for other proofs of this fact. Again, the statement is, if $K$ is a compact operator on a Hilbert space ...
3
votes
1answer
244 views

Spectral theorem of compact operators in Hilbert space

I am reading the following theorem from my lecture notes (English translation of German text). But I don't understand exactly what is meant from this theorem and the proof. Theorem. Let $H$ ...
1
vote
1answer
126 views

Spectral radius of an operator .

I would like to know the spectral radius of the operator $T_k$ from $C[0,1] \to C[0,1]$ : $$T_k x (t)= \int_0^1 k(t,s) x(s) ds$$ where $k(x,y)\colon [0,1]^2 \to \mathbb C$ is continuous. And ...
3
votes
1answer
324 views

Eigenvalues integral operator - general case

Let $T$ be an integral operator on $L^2([0,1])$, such that: $$ (Tf)(x) = \int_0^1K(x,y)f(y)dy, $$ with $K(x,y): [0,1]^2 \rightarrow \mathbb{R}$ continuous and $K(x,y) = K(y,x)$, $K(x,y)\geq0$ $ ...
3
votes
1answer
449 views

Compactness and spectrum of integral operator

Show that the operator $C: L^2([0,1]) \rightarrow L^2([0,1])$ defined by $$Cf(x) = \int_0^x\int_1^tf(s)dsdt$$ is compact and determine its spectrum. Im not sure how to find the spectrum when we are ...
3
votes
1answer
123 views

Eigenvalues of Hilbert-Schmith operator

I am having trouble determining the eigenvalues and eigenvectors of the operator $Kv(x)= \int_0^1((x+t)v(t)dt$, where the kernel is $k=x+t$. I have tried to solve the equation $Kv(x)=\lambda v(x)$, ...