Basis-free formulation of Jordan normal form theorem Is there a basis-free formulation of Jordan normal form theorem?
From some search I did in Google, the answer is apparently yes. But I didn't find any article that I could understand. (I've only taken two semester course in linear algebra.)
My curiosity comes from the question whether the theorem can be generalized to infinite dimensional situation. If it's a separable Hilbert space, we can still represent the linear operator as a matrix, but does the theorem remain true?
In case of non-separable space, I think there's no way to put the linear operator in matrix form. So we need to find a basis-free formulation.
Wikipedia says that there is an analogue of Jordan normal form theorem for compact operators in Banach space. What is this analogous result?
 A: A basis-free form of the Jordan normal form theorem is as follows. Let $T : V \to V$ be a linear operator on a finite-dimensional vector space $V$ over an algebraically closed field $k$. Then we can write $T = D + N$ where $D$ is diagonalizable (this is a basis-free condition), $N$ is nilpotent, and $DN = ND$. 
A way of saying "diagonalizable" which might be more clearly basis-free is semisimple. The reason is that any linear operator $T$ on a vector space $V$ gives $V$ the structure of a $k[x]$-module where $x$ acts by $T$, and this module is semisimple if and only if $T$ is diagonalizable. 

As Martin says, this statement is badly false in infinite dimensions. The Jordan normal form theorem implies that $T$ has an eigenvector, and this is false in general. For example, let $e_1, e_2, ... $ be an orthonormal basis of an infinite-dimensional separable Hilbert space $H$ and let $T : H \to H$ be defined by
$$T e_i = e_{i+1}.$$
An eigenvector for $T$ is a vector $v = \sum c_n e_n$ such 
$$T v = \sum c_n e_{n+1} = \lambda v = \sum \lambda c_n e_n.$$
This gives $c_1 = 0$ and $c_n = \lambda c_{n+1}$. If $\lambda = 0$ then $c_n = 0$ for all $n$, and if $\lambda \neq 0$ then again $c_n = 0$ for all $n$. 
Generalizing to infinite dimensions is quite nontrivial, depending on how strong you want your results to be; see spectral theory for a further discussion. The nicest statements are available for compact self-adjoint operators on Hilbert spaces. 
A: It is true that you can think of an operator on a Hilbert space as an "infinite matrix". But little to nothing can be obtained from that. So much, that it doesn't make a difference whether your space is separable or not. 
Regarding your concrete question, if you mean the Jordan form in the same sense as in matrix theory (the operator being a direct sum of scalar multiples of the identity plus nilpotents) the answer is that no such generalization exists. Because such an operator would have eigenvalues, and in any infinite-dimensional Hilbert space you have operators with no eigenvalues. 
