Is it possible to find a basis for a non-compact operator?

Question

Is it possible to find a basis for a non-compact operator?

I was thinking of applying the spectral theorem to non-compact operators.

Answer 1

As this wiki page notes, there is no guarantee that a (self-adjoint) non-compact operator has any eigenvectors, let alone enough to form a basis (orthonormal or otherwise). In particular, the multiplication operator $A:L^2[0,1] \to L^2[0,1]$ defined by $[A\phi](t) = t\phi(t)$ has no eigenvectors.

Answer 2

The spectral theorem for self-adjoint (or more generally hermitian) operators in Hilbert spaces is a standard topic in functional analysis. It will occupy a prominent place in almost any functional analysis textbook. But you can't use eigenvectors. Rather, standard formulation are in terms of projection-valued measures (resolutions of the identity), or the continuous functional calculus.