Editing Spectrum (functional analysis) (section)

===Relation to eigenvalues===

If <math>\lambda</math> is an eigenvalue of <math>T</math>, then the operator <math>T-\lambda I</math> is not one-to-one, and therefore its inverse <math>(T-\lambda I)^{-1}</math> is not defined.  However, the converse statement is not true: the operator <math>T - \lambda I</math> may not have an inverse, even if <math>\lambda</math> is not an eigenvalue. Thus the spectrum of an operator always contains all its eigenvalues, but is not limited to them.

For example, consider the Hilbert space <math>\ell^2(\Z)</math>, that consists of all [[Sequence#Finite and infinite|bi-infinite sequences]] of real numbers
:<math>v = (\ldots, v_{-2},v_{-1},v_0,v_1,v_2,\ldots)</math>
that have a finite sum of squares <math display="inline">\sum_{i=-\infty}^{+\infty} v_i^2</math>.  The [[bilateral shift]] operator <math>T</math> simply displaces every element of the sequence by one position; namely if <math>u = T(v)</math> then <math>u_i = v_{i-1}</math> for every integer <math>i</math>.  The eigenvalue equation <math>T(v) = \lambda v</math> has no nonzero solution in this space, since it implies that all the values <math>v_i</math> have the same absolute value (if <math> \vert \lambda \vert = 1</math>) or are a geometric progression (if <math> \vert \lambda \vert \neq 1</math>); either way, the sum of their squares would not be finite.  However, the operator <math>T-\lambda I</math> is not invertible if <math>|\lambda| = 1</math>. For example, the sequence <math>u</math> such that <math>u_i = 1/(|i|+1)</math> is in <math>\ell^2(\Z)</math>; but there is no sequence <math>v</math> in <math>\ell^2(\Z)</math> such that <math>(T-I)v = u</math> (that is, <math>v_{i-1} = u_i + v_i</math> for all <math>i</math>).