Spectral theorem from Jordan decomposition in infinite dimensions

Question

The finite-dimensional spectral theorem is a simple special case of the finite-dimensional Jordan decomposition. We also have various infinite-dimensional generalizations of the spectral theorem; can these be seen as coming from an infinite-dimensional version of Jordan decomposition?

Answer 1

There are also infinite dimensional generalizations (say for bounded operators on Banach spaces) of the Jordan decomposition, but the situation is of course more complicated. I'm certainly not well informed on the state of the art. But with a sufficiently strong hypothesis, like compactness, one can achieve a pretty complete statement. The following result is already in §80 of Riesz & Nagy's classic Functional Analysis (Ungar, 1955):

For a compact operator $K$ on a Hilbert space, the spectrum is discrete (accumulating at $0$) and the invariant subspace corresponding to each non-zero eigenvalue is finite dimensional (hence, $K$ restricted to each such invariant subspace has the usual Jordan decomposition).

Another reference for a discussion of this kind is in Chapter VII of Dunford & Schwartz's Linear Operators, Part I: General Theory (Interscience, 1958).

Results for more general bounded or unbounded operators (also just looking at compact operators with infinite dimensional subspaces corresponding to the zero eigenvalue) are obstructed by the related Invariant Subspace Problem, which is still open. It asks whether any bounded operator on a Banach space has a non-trivial closed invariant subspace. If such an operator without a non-trivial closed invariant subspace existed, it would behave very differently from any finite dimensional operator (of size larger than $1\times 1$), which is guaranteed to have a $1$-dimensional invariant subspace in each Jordan block. Even under more restrictive hypotheses, the examples of shift operators seem to confound reasonable notions of a nilpotent operator such that a reasonable class of operators could always be decomposed as $S + N$, where $S$ is semi-simple (diagonalizable, essentially) and $N$ nilpotent. See for instance the discussion on this math.SE question.

Answer 2

An important consequence of the Jordan form in $C^n$ is that given a polynomial $P$ with complex coefficients and a matrix $A$, the value of $P(A)$ depends on $P$ and its derivatives only at the eigenvalues of $A$. For example,

$$P\left( \left[ \begin{array}{cccc} \lambda & 1 & 0 & 0 \\ & \lambda & 1 & 0 \\ & & \lambda & 1 \\ & & & \lambda \end{array}% \right] \right) =\left[ \begin{array}{cccc} P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 & P^{\prime \prime \prime }\left( \lambda \right) /6 \\ & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) & P^{\prime \prime }\left( \lambda \right) /2 \\ & & P\left( \lambda \right) & P^{\prime }\left( \lambda \right) \\ & & & P\left( \lambda \right) \end{array}% \right] $$

In particular, in the finite-dimensional case if $f:D \rightarrow C$ is an analytic function on a domain containing the spectrum of $A$ then one may uniquely define $$f(A)=P(A),$$ where $P$ is any polynomial which interpolates $f$ and its derivatives at the eigenvalues of $A$ $$P^{(n)}(\lambda_k)=f^{(n)}(\lambda_k),$$ where for each eigenvalue $\lambda_k$ the order of the derivative $n$ is less than or equal to the longest chain of 1's in the Jordan block.

This definition has nice properties, such as $f(g(A))=(f\circ g)(A)$, and it agrees with the power series definition for the matrix exponential, ect.

Indeed, this is just the finite-dimensional version of the usual definition of $f(A)$ under the so-called "Dunford calculus," https://en.wikipedia.org/wiki/Holomorphic_functional_calculus which uses the Cauchy integral formula and applies in infinite-dimensions.

If one views the purpose of the Jordan form in finite-dimensions as the creation of an analytic calculus, then the Dunford calculus extends this purpose to infinite dimensions. So one might argue that the Dunford calculus essentially is the generalization of the Jordan form.

Now the spectral theorem allows one to define the Borel functional calculus, computing $f(A)$ for Borel measureable $f:\sigma(A)\rightarrow R$ and self-adjoint $A$. It's not obvious to me how could obtain the Borel calculus from a holomorphic calculus in infinite dimensions, in part because the spectrum of $A$ need no longer be discrete.