 Solution

Work over a complex Hilbert space; for a real space one first complexifies to discuss a complex spectrum. A bounded linear operator $L$ is self-adjoint when $L=L^{\ast }$, where the adjoint operator is defined byEquivalently, $⟨Lh,k⟩=⟨h,Lk⟩$ for every pair of vectors.

The spectrum of a bounded operator isIts complement is the resolvent set. On a Banach space, a bounded bijective operator has a bounded inverse by the bounded inverse theorem, so failure of invertibility is equivalent to failure of bijectivity. An eigenvalue is one possible spectral point, but an injective operator that is not onto can also contribute to the spectrum.