Editing Spectral theorem (section)

===Possible absence of eigenvectors===

The next generalization we consider is that of [[Self-adjoint_operator#Bounded_self-adjoint_operators|bounded self-adjoint operators]] on a Hilbert space. Such operators may have no eigenvectors: for instance let {{math|''A''}} be the operator of multiplication by {{math|''t''}} on <math>L^2([0,1])</math>, that is,<ref>{{harvnb|Hall|2013}} Section 6.1</ref>
<math display="block"> [A f](t) = t f(t). </math>

This operator does not have any eigenvectors ''in'' <math>L^2([0,1])</math>, though it does have eigenvectors in a larger space. Namely the [[Distribution (mathematics)|distribution]] <math>f(t)=\delta(t-t_0)</math>, where <math>\delta</math> is the [[Dirac delta function]], is an eigenvector when construed in an appropriate sense. The Dirac delta function is however not a function in the classical sense and does not lie in the Hilbert space {{math|''L''<sup>2</sup>[0, 1]}}. Thus, the delta-functions are "generalized eigenvectors" of <math>A</math> but not eigenvectors in the usual sense.