Editing Spectral theorem (section)

== Compact self-adjoint operators ==
{{see also|Compact operator on Hilbert space#Spectral theorem}}
In the more general setting of Hilbert spaces, which may have an infinite dimension, the statement of the spectral theorem for [[compact operator|compact]] [[self-adjoint operators]] is virtually the same as in the finite-dimensional case.

{{math theorem | math_statement =Suppose {{math|''A''}} is a compact self-adjoint operator on a (real or complex) Hilbert space {{math|''V''}}. Then there is an [[orthonormal basis]] of {{math|''V''}} consisting of eigenvectors of {{math|''A''}}. Each eigenvalue is real.}}

As for Hermitian matrices, the key point is to prove the existence of at least one nonzero eigenvector. One cannot rely on determinants to show existence of eigenvalues, but one can use a maximization argument analogous to the variational characterization of eigenvalues. 

If the compactness assumption is removed, then it is ''not'' true that every self-adjoint operator has eigenvectors. For example, the multiplication operator <math>M_{x}</math> on <math>L^2([0,1])</math> which takes each <math>\psi(x) \in L^2([0,1])</math> to <math>x\psi(x)</math> is bounded and self-adjoint, but has no eigenvectors. However, its spectrum, suitably defined, is still equal to <math>[0,1]</math>, see [[Spectrum_(functional_analysis)#Spectrum_of_a_bounded_operator| spectrum of bounded operator]].