Let H be an infinite dimensional non-separable real Hilbert space and T:H->H be a linear and bounded operator. Let s be an eigenvalue of T. The set Es(T)={x/T(x)=sx} is a closed subspace of H.
In the above situation, how can be represented the orthogonal complement of a subspace of the form Es(T), using only other subspaces of the form Es(T)?
What happens if T is self-adjoint or compact? Or self-adjoint and compact simultaneous? What happens in case of linear integral operators from L2 to L2?
Remark It is known that: If T is self-adjoint and a and b are different eigenvalues of T, then Ea(T) and Eb(T) are orthogonal.
In general this is not possible. The operator may have an isolated eigenvalue and no other one in the rest of the spectrum (2 x identity (direct sum) forward shift, for example) and the orthocomplement of the eigenspace is the space of the forward shift which has no description in terms of eigenspaces.
It may be possible in the case when the spectrum is generated by eigenspaces (the Cowen - Douglas operators) but I believe it will work only when the eigenvalues are isolated, like in the case of a compact selfadjoint.
The reason I believe it may not be possible is because I am thinking at the case of the backward shift. The space is generated by the eigenspaces but it is in fact generated by any countable set of eigenspaces if the set of corresponding eigenvalues has a cluster point inside the open unit disk. If it does not have a cluster point inside but only on the boundary that I don't think it generates the space anymore.
In short, you either get too much or too little, it does not seem possible to et the orthocomplement of one particular eigenspace.
This may be a solved problem though, Kato's books will be the right place to look, and also Herrero's and some papers of Constantin Apostol (his joint papers with Bernie Morrel and with Clancey).
Thanks Gabriel for your advised answer and for the mentioned sources! I will search for information about.
In the case of a compact self adjoint operator T, the spectrum is a real numbers set, is identical with the set of eigenvalues, is countable(so contains only isolated points, as you specified) and bounded by m=inf{ (Tx,x) with //x//=1} and M=sup{ (Tx,x) with //x//=1}, where ( , ) is the inner product in H.
In this case, if the space H would be separable, I think that H may be represented as direct sum off all eigenspaces. But I do not need separable Hilbert spaces, I need properties for L2 or for hilbertian Sobolev spaces.
I can adjust my linear integral operator, imposing conditions on its kernel, and it would not be a too restrictive situation.
Also, connections between Rg(T) and eigenspaces of T may be useful for me!
Dear Dinu Teodorescu
If T is a self-adjoint operator then by the projection theorem H=N(T)\oplus N(T)^\perp=N(T)\oplus \overline{R(T)}. Now, if we add the condition that T is compact, then by the spectral theorem, H=N(T)\oplus[\oplus_{k\in N} N(T-\lambda_k)]. This implies that for every p\in N, N(T-\lambda_p)^\perp=N(T)\oplus[\oplus_{k\in N\setminus{p}}N(T-\lambda_k)].
Best, Salvador
Salvador Sánchez-Perales
Thanks Salvador! Would you, please, sent the decompositions of H, without using LaTeX in the text!
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