Editing Jordan normal form (section)

=== The finite-dimensional case ===

In the finite-dimensional case, ''σ''(''T'') = {''λ''<sub>''i''</sub>} is a finite discrete set in the complex plane. Let ''e''<sub>''i''</sub> be the function that is 1 in some open neighborhood of ''λ''<sub>''i''</sub> and 0 elsewhere. By property 3 of the functional calculus, the operator

:<math>e_i(T)</math>

is a projection. Moreover, let ''ν<sub>i</sub>'' be the index of ''λ''<sub>''i''</sub> and

:<math>f(z)= (z - \lambda_i)^{\nu_i}.</math>

The spectral mapping theorem tells us

:<math> f(T) e_i (T) = (T - \lambda_i)^{\nu_i} e_i (T)</math>

has spectrum {0}. By property 1, ''f''(''T'') can be directly computed in the Jordan form, and by inspection, we see that the operator ''f''(''T'')''e<sub>i</sub>''(''T'') is the zero matrix.

By property 3, ''f''(''T'') ''e''<sub>''i''</sub>(''T'') = ''e''<sub>''i''</sub>(''T'') ''f''(''T''). So ''e''<sub>''i''</sub>(''T'') is precisely the projection onto the subspace

:<math>\operatorname{Ran} e_i (T) = \ker(T - \lambda_i)^{\nu_i}.</math>

The relation

:<math>\sum_i e_i = 1</math>

implies

:<math>\mathbb{C}^n = \bigoplus_i \; \operatorname{Ran} e_i (T) = \bigoplus_i \ker(T - \lambda_i)^{\nu_i}</math>

where the index ''i'' runs through the distinct eigenvalues of ''T''. This is the invariant subspace decomposition

:<math>\mathbb{C}^n = \bigoplus_i Y_i</math>

given in a previous section. Each ''e<sub>i</sub>''(''T'') is the projection onto the subspace spanned by the Jordan chains corresponding to ''λ''<sub>''i''</sub> and along the subspaces spanned by the Jordan chains corresponding to v<sub>''j''</sub> for ''j'' ≠ ''i''. In other words, ''e<sub>i</sub>''(''T'') = ''P''(''λ''<sub>''i''</sub>;''T''). This explicit identification of the operators ''e<sub>i</sub>''(''T'') in turn gives an explicit form of holomorphic functional calculus for matrices:

:For all ''f'' ∈ Hol(''T''),

:<math>f(T) = \sum_{\lambda_i \in \sigma(T)} \sum_{k = 0}^{\nu_i -1} \frac{f^{(k)}}{k!} (T - \lambda_i)^k e_i (T).</math>

Notice that the expression of ''f''(''T'') is a finite sum because, on each neighborhood of v<sub>''i''</sub>, we have chosen the [[Taylor series]] expansion of ''f'' centered at v<sub>''i''</sub>.