Editing Jordan normal form (section)

=== Poles of an operator ===

Let ''T'' be a bounded operator ''λ'' be an isolated point of ''σ''(''T''). (As stated above, when ''T'' is compact, every point in its spectrum is an isolated point, except possibly the limit point 0.)

The point ''λ'' is called a '''pole''' of operator ''T'' with order ''ν'' if the [[Resolvent formalism|resolvent]] function ''R''<sub>''T''</sub> defined by

:<math> R_T(\lambda) = (\lambda - T)^{-1}</math>

has a [[pole (complex analysis)|pole]] of order ''ν'' at ''λ''.

We will show that, in the finite-dimensional case, the order of an eigenvalue coincides with its index. The result also holds for compact operators.

Consider the annular region ''A'' centered at the eigenvalue ''λ'' with sufficiently small radius ''ε'' such that the intersection of the open disc ''B<sub>ε</sub>''(''λ'') and ''σ''(''T'') is {''λ''}. The resolvent function ''R''<sub>''T''</sub> is holomorphic on ''A''.
Extending a result from classical function theory, ''R''<sub>''T''</sub> has a [[Laurent series]] representation on ''A'':

:<math>R_T(z) = \sum_{-\infty}^\infty a_m (\lambda - z)^m</math>

where

:<math>a_{-m} = - \frac{1}{2 \pi i} \int_C (\lambda - z) ^{m-1} (z - T)^{-1} d z</math> and ''C'' is a small circle centered at&nbsp;''λ''.

By the previous discussion on the functional calculus,

:<math> a_{-m} = -(\lambda - T)^{m-1} e_\lambda (T)</math> where <math> e_\lambda</math> is 1 on <math> B_\varepsilon(\lambda)</math> and 0 elsewhere.

But we have shown that the smallest positive integer ''m'' such that

:<math>a_{-m} \neq 0</math> and <math>a_{-l} = 0 \; \; \forall \; l \geq m</math>

is precisely the index of ''λ'', ''ν''(''λ''). In other words, the function ''R''<sub>''T''</sub> has a pole of order ''ν''(''λ'') at ''λ''.