# Why we still diagonalize compact operators even tho we lack invertibility.

We know that any compact symmetric operator on a Hilbert space, has a orthogonal base of eigenvectors. But we also know that $0$ is in the spectrum if $X$ is infinite dimensional, which makes the operator non-invertible(in particular they are never surjective).

Question;

Why we would care to diagonalise in infinite dimensional at all since never get as good info as in finite dimension (non-zero eigenvalues on the diagonal).

Diagonalisation of a compact operator $C$ in infinite dimensions seems more related to the invertabiity(or solvability) of Iλ−C (a fredholm equation of 2nd kind) for compact operator C rather then properties of C itself as a linear map which often is the object of intrest in finite dimension i.e linear algebra. That might be a reason for my confusion. From a historcial point of view we seem to care about solvability of Iλ−C and qualitive properties of solutions rather then C itself as a map from one vector space into another,in some sense. So when we "diagonalize" we really just cheack solvabilty of Iλ−C, we are not looking for a nice bijection(the full diagonal matrix) of the vectorspaces as we do in linear algebra.

Or simply the MAIN reason we started to study the spetrum of an operator $C$ was to determine if $I \lambda - C$ is solvable.

Am I on to something or just way off?

• You mean "compact diagonal operators" in the title? Certainly there are plenty of invertible diagonal operators (e.g. the identity operator), which are not compact. – Nate Eldredge Nov 18 '15 at 14:52
• @NateEldredge yes and thanks – user1 Nov 18 '15 at 14:54
• I'm not entirely clear what you are asking - why would this fact make diagonalization less useful? But it may help to consider that a typical example of a compact diagonal operator is one where the $n$th diagonal entry is $1/n$. None of the eigenvalues is 0 (this operator is injective) but they are getting very close. This basically has to happen: the diagonal entries must be a sequence approaching 0. – Nate Eldredge Nov 18 '15 at 14:59
• @NateEldredge I guess I would like diagonalisation to lead to some kind of simple bijection as in finite dimension. I cant see what we gaining in this case? – user1 Nov 18 '15 at 15:06
• What bijection are you talking about? – Nate Eldredge Nov 18 '15 at 15:08

For any bounded selfadjoint operator $T$ on a Hilbert space $X$, you can reduce to the case where $T$ is injective because $X=\mathcal{N}(T)\oplus\overline{\mathcal{R}(T)}$ and both of these closed subspaces are invariant under $T$. That allows you to reduce to looking at $T$ on the Hilbert space $Y=\overline{\mathcal{R}(T)}$, and $T$ remains selfadjoint on this invariant subspace.

The operator $T : Y\rightarrow Y$ has dense range and is injective. If the range of $T$ is closed, then $T$ is a topological isomorphism, which means that the unit ball of $Y$ is compact, leading to the conclusion that $Y$ is finite-dimensional. Otherwise, the range of $T$ cannot be closed, which means that $0 \in\sigma(T)$.

• I been trying to understand $X=\mathcal{N}(T)\oplus\overline{\mathcal{R}(T)}$. But cant seem to manage :/ – user1 Nov 19 '15 at 13:28
• @User2313 : For any bounded operator $T$, you have $(Tx,y)=(x,T^{\star}y)$. Therefore $y \perp \mathcal{R}(T)$ iff $(x,T^{\star}y)=0$ for all $x$, which holds iff $T^{\star}y=0$. So $\mathcal{R}(T)^{\perp}=\mathcal{N}(T^{\star})$. So $\overline{\mathcal{R}(T)}=\mathcal{N}(T^{\star})^{\perp}$. – DisintegratingByParts Nov 19 '15 at 15:58
• Dont you need to assume compactness of $T$ in the second paragraph? – user1 Nov 20 '15 at 5:43
• @User2313 : The compactness comes in at the point of noting that $Y$ is finite-dimensional if $T$ is compact and has a closed range. – DisintegratingByParts Nov 20 '15 at 7:17
• Very nice, do you have some motivation for the reason why still diagonalise? As you can see in comments this is somthing which puzzles me and maybe some clarification what I mean by "good info". – user1 Nov 21 '15 at 13:30

You asked for intuition so I am going to allow myself to play a little loose...

Let $U$ and $V$ be infinite vector spaces with a countable bases $\{e_i\}$ and $\{f_i\}$. Suppose that a linear map $T:U\rightarrow V$ is diagonalisable such that

$$T\left(\sum_i\alpha_i e_i\right)=\sum_i\alpha_i \lambda_i f_i.$$

If none of the $\lambda_i$ are zero, and so $T$ invertible, then the range of $T$ will be in a closed ball of infinite dimension and so $T$ cannot be compact.

• This is nice but it does not adress what Im asking. – user1 Nov 18 '15 at 17:06
• Perhaps your title and question at odds? Does this not answer your title? – JP McCarthy Nov 18 '15 at 17:07
• It answers the title yes, but not the questions in the question. – user1 Nov 18 '15 at 17:09
• Your right, Im thinking of a better title. Sry about that. – user1 Nov 18 '15 at 17:42
• I'm intrigued by your intuition, but it was quite a lot since I studied topology so I don't exactly remember... doesn't some Tychonoff theorem says that even an infinite dimensional space can be compact? – Dac0 Nov 26 '15 at 18:44