# Jordan form of compact operator [closed]

Let $X$ be Banach space over a field $\mathbb{C}$. Consider the Banach space $L_c$ of compact operators in $X$. Let $A^0\in L_c$ be fixed and $\lambda^0\neq 0$ his eigenvalue with algebraic multiplicity $m$ which is associated with Jordan form $J_{\overline k}(\lambda^0)$.

In the canonical basis of the subspace that corresponding to eigenvalue $\lambda^0$, operator $A^0$ has the form $$A^0=\begin{pmatrix}J_{\overline k}(\lambda^0)&0 \\\ 0 & A_{\textbf{2,2}}^0 \end{pmatrix}.$$

Questions.

1. Can I speak about canonical basis in Banach space or better use a Hilbert space?
2. Is it natural to consider the Jordan form of compact operator on Banach or Hilbert space?
3. Does any non-zero eigenvalue of a compact operator have finite multiplicity?

21.11. Operator $A^0$ such that it has non-zero eigenvalues.

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## closed as too localized by Bill Johnson, Yemon Choi, Andrés Caicedo, Mark Sapir, Ryan BudneyNov 22 '11 at 18:46

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Ringrose's little book "Compact non-self-adjoint operators" has a good exposition of how things work in the Hilbert space case. The basics of the theory for compact operators on general Banach spaces is in Rudin's Functional Analysis, if I recall correctly. – Yemon Choi Nov 20 '11 at 23:32
I'm also not quite sure what you're looking for with your questions; perhaps I have misunderstood something. We know that for the non-zero eigenvalues, the eigenspaces are finite-dimensional; so all the theory for the finite-dimensional case applies on each such eigenspace. On the other hand, one can have quasinilpotent compact operators on Banach spaces and there all bets are off unless one knows more about the invariant subspaces of the operator (like the Volterra case mentioned in on eof the answers) – Yemon Choi Nov 20 '11 at 23:36
What do you mean by canonical basis? There is certainly more than one choice of eigenbasis if you just have a two-dimensional eigenspace, for example. – Christopher A. Wong Nov 21 '11 at 10:45

I'm not entirely sure what you mean by 1 and 2, but here is what can be said in general about compact operators:

The spectrum of a compact operator $A$ on $X$ consists of 0 (for infinite-dimensional $X$) and at most countably many eigenvalues with finite multiplicity (this answers question 3), and for such an eigenvalue $\lambda\ne 0$ there is the decomposition $X=N\oplus R$ (apparently this is what you mean by "canonical basis") into the generalized eigenspace (the kernel of $(\lambda-A)^k$ for suitable $k$) and the remainder space, on which $\lambda-A$ is an isomorphism. This decomposition is respected by $A$ (indicated by the off-diagonal zeros in your matrix).

All this works in any Banach space (does that answer question 1?), but the problem is that an operator might not have any nonzero eigenvalues. The situation is much better if $X$ is a complex Hilbert space and $A$ is normal (something similar for non-normal operators probably can be said but the textbook I just grabbed does not say anything about this). In this case there always exists a nonzero eigenvalue unless $A=0$, and this yields the spectral theorem which says that $X$ can be decomposed into a direct sum of kernel and eigenspaces. Thus you might argue that considering eigenvalues is more natural on Hilbert spaces than it is on Banach spaces; at least it is more useful.

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I don't know what you expect to say about nonnormal compact operators on a Hilbert space. They too can have no nonzero eigenvalues. A useful example might be $T$ defined on $\ell^2$ by $(T x)_n = x_{n+1}/n$. Note that in this case the union of the kernels of powers of $T$ is dense in $\ell^2$. – Robert Israel Nov 20 '11 at 20:52
One fact about compact operators that is true in Hilbert space but not in all Banach spaces is that compact operators can be approximated (in norm) by finite-rank operators (see <en.wikipedia.org/wiki/Approximation_property>). – Robert Israel Nov 20 '11 at 21:03
I choose the operator $A^0$ such that it has non-zero eigenvalues. – Alexander Nov 21 '11 at 7:58
The point is that for a nonzero eigenvalue, its eigenspace can be separated, and on the remaining space the operator no longer has this eigenvalue. A countable repetition of this process yields a decomposition into the various eigenspaces and a remaining space on which there are no nonzero eigenvalues, and at this point you are stuck. So it doesn't help you too much that $A^0$ just has one eigenvalue. – Florian Nov 21 '11 at 17:14

In addition to Florian's answer, there exists a much more complicated theory of Jordan's form for some classes of Volterra operators on Hilbert space, that is compact operators without nonzero eigenvalues. See

I. Gohberg and M. G. Krein. Theory and applications of Volterra operators in Hilbert space. Providence, R.I.: American Mathematical Society (AMS) (1970).

M. S. Brodskij, Triangular and Jordan representations of linear operators. Providence, R.I.: American Mathematical Society (1971).

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