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?




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



Jordan form of compact operator
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.


