Editing Jordan normal form (section)

== Generalized eigenvectors ==
{{main|Generalized eigenvector}}
Given an eigenvalue ''λ'', every corresponding Jordan block gives rise to a '''[[Jordan chain]]''' of linearly independent vectors ''p<sub>i</sub>, i'' = 1, ..., ''b'', where ''b'' is the size of the Jordan block. The '''generator''', or '''lead vector''', ''p<sub>b</sub>'' of the chain is a generalized eigenvector such that {{tmath|1=(A - \lambda I)^b p_b = 0}}. The vector {{tmath|1=p_1 =  (A - \lambda I)^{b-1} p_b}} is an ordinary eigenvector corresponding to ''λ''. In general, ''p''<sub>''i''</sub> is a preimage of ''p''<sub>''i''−1</sub> under {{tmath|1=A - \lambda I}}. So the lead vector generates the chain via multiplication by {{tmath|1=A - \lambda I}}.<ref>{{harvtxt|Bronson|1970|pp=189,194}}</ref><ref name="Holt 2009 9" /> Therefore, the statement that every square matrix ''A'' can be put in Jordan normal form is equivalent to the claim that the underlying vector space has a basis composed of Jordan chains.

=== 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.

=== Uniqueness ===

It can be shown that the Jordan normal form of a given matrix ''A'' is unique up to the order of the Jordan blocks.

Knowing the algebraic and geometric multiplicities of the eigenvalues is not sufficient to determine the Jordan normal form of ''A''. Assuming the algebraic multiplicity ''m''(''λ'') of an eigenvalue ''λ'' is known, the structure of the Jordan form can be ascertained by analyzing the ranks of the powers {{math|1=(''A'' − ''λI'')<sup>''m''(''λ'')</sup>}}. To see this, suppose an ''n'' × ''n'' matrix ''A'' has only one eigenvalue ''λ''. So ''m''(''λ'') = ''n''. The smallest integer ''k''<sub>1</sub> such that

:<math>(A - \lambda I)^{k_1} = 0</math>

is the size of the largest Jordan block in the Jordan form of ''A''. (This number ''k''<sub>1</sub> is also called the '''index''' of ''λ''. See discussion in a following section.) The rank of

:<math>(A - \lambda I)^{k_1 - 1}</math>

is the number of Jordan blocks of size ''k''<sub>1</sub>. Similarly, the rank of

:<math>(A - \lambda I)^{k_1 - 2}</math>

is twice the number of Jordan blocks of size ''k''<sub>1</sub> plus the number of Jordan blocks of size ''k''<sub>1</sub> − 1. The general case is similar.

This can be used to show the uniqueness of the Jordan form. Let ''J''<sub>1</sub> and ''J''<sub>2</sub> be two Jordan normal forms of ''A''. Then ''J''<sub>1</sub> and ''J''<sub>2</sub> are similar and have the same spectrum, including algebraic multiplicities of the eigenvalues. The procedure outlined in the previous paragraph can be used to determine the structure of these matrices. Since the rank of a matrix is preserved by similarity transformation, there is a bijection between the Jordan blocks of ''J''<sub>1</sub> and ''J''<sub>2</sub>. This proves the uniqueness part of the statement.