Editing Jordan normal form (section)

=== A proof ===
We give a [[proof by induction]] that any complex-valued square matrix ''A'' may be put in Jordan normal form. Since the underlying vector space can be shown<ref>Roe Goodman and Nolan R. Wallach, ''Representations and Invariants of Classical Groups'', Cambridge UP 1998, Appendix B.1.</ref> to be the direct sum of [[invariant subspace]]s associated with the eigenvalues, ''A'' can be assumed to have just one eigenvalue ''λ''. The 1 × 1 case is trivial. Let ''A'' be an ''n'' × ''n'' matrix. The [[range of a function|range]] of {{tmath|1=A - \lambda I}}, denoted by {{tmath|1=\operatorname{Ran}(A - \lambda I)}}, is an invariant subspace of ''A''. Also, since ''λ'' is an eigenvalue of ''A'', the dimension of {{tmath|1=\operatorname{Ran}(A - \lambda I) }}, ''r'', is strictly less than ''n'', so, by the inductive hypothesis, {{tmath|1=\operatorname{Ran}(A - \lambda I) }} has a [[basis (linear algebra)|basis]] {''p''<sub>1</sub>, ..., ''p''<sub>''r''</sub>} composed of Jordan chains.
 
Next consider the [[kernel (linear algebra)|kernel]], that is, the [[linear subspace|subspace]] {{tmath|1=\ker(A - \lambda I)}}. If

:<math>\operatorname{Ran}(A - \lambda I) \cap \ker(A - \lambda I) = \{0\},</math>

the desired result follows immediately from the [[rank–nullity theorem]]. (This would be the case, for example, if ''A'' were [[Hermitian matrix|Hermitian]].)

Otherwise, if

:<math>Q = \operatorname{Ran}(A - \lambda I) \cap \ker(A - \lambda I) \neq \{0\},</math>

let the dimension of ''Q'' be {{math|1=''s'' ≤ ''r''}}. Each vector in ''Q'' is an eigenvector, so {{tmath|1=\operatorname{Ran}(A - \lambda I)}} must contain ''s'' Jordan chains corresponding to ''s'' linearly independent eigenvectors. Therefore the basis {''p''<sub>1</sub>, ..., ''p''<sub>''r''</sub>} must contain ''s'' vectors, say {''p''<sub>1</sub>, ..., ''p<sub>s</sub>''}, that are lead vectors of these Jordan chains. We can "extend the chains" by taking the preimages of these lead vectors. (This is the key step.) Let ''q''<sub>''i''</sub> be such that

:<math>\; (A - \lambda I) q_i = p_i \mbox{ for } i = 1, \ldots,s.</math>

Finally, we can pick any basis for

:<math>\ker(A - \lambda I) / Q</math>

and then lift to vectors  {''z''<sub>1</sub>, ..., ''z''<sub>''t''</sub>} in {{tmath|1=\ker(A - \lambda I)}}. Each ''z''<sub>''i''</sub> forms a Jordan chain of length 1. We just need to show that the union of {''p''<sub>1</sub>, ..., ''p''<sub>''r''</sub>}, {''z''<sub>1</sub>, ..., ''z''<sub>''t''</sub>}, and {''q''<sub>1</sub>, ..., ''q<sub>s</sub>''} forms a basis for the vector space. 

By the rank-nullity theorem, {{tmath|1=\dim(\ker(A - \lambda I)))=n-r}}, so {{tmath|1=t=n-r-s}}, and so the number of vectors in the potential basis is equal to n. To show linear independence, suppose some linear combination of the vectors is 0. Applying {{tmath|1=A - \lambda I,}} we get some linear combination of ''p''<sub>i</sub>, with the ''q<sub>i</sub>'' becoming lead vectors among the ''p''<sub>i.</sub> From linear indepence of ''p''<sub>i,</sub> it follows that the coefficients of the vectors ''q<sub>i</sub>''  must be zero. Furthermore, no non-trivial linear combination of the ''z<sub>i</sub>'' can equal a linear combination of ''p''<sub>''i''</sub>, because then it would belong to {{tmath|1=\operatorname{Ran}(A - \lambda I)}} and thus {{mvar|Q}}, which is impossible by the construction of ''z<sub>i</sub>''. Therefore the coefficients of the ''z<sub>i</sub>'' will also be 0. This leaves in the original linear combination just the ''p<sub>i</sub>'' terms, which are assumed to be linearly independent, and so their coefficients must be zero too. We have found a basis composed of Jordan chains, and this shows ''A'' can be put in Jordan normal form.