# 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?

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I believe it is not an elementary problem. The Jordan canonical form is actually a statement about the existence of invariant subspaces for the operator. This is the approach of Herstein in his book on Algebra. In infinite dimension, you no longer have coordinates, but the idea of invariant subspaces survives. However, you need spectral theory and functional calculus, as stated in en.wikipedia.org/wiki/… –  Siminore Apr 28 '12 at 12:38
In non-separable Hilbert space, we can use matrices with uncountably many rows and columns. –  GEdgar Apr 28 '12 at 12:41
Do such matrices have "diagonal"? –  makela Apr 28 '12 at 12:45

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.

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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.

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