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The Generalized Eigenvectors of a Matrix
and their Linear Indepedence

Let A be an n×n matrix and let λ be an eigenvalue of A. Let 0 denote the vector of zeroes. A nonzero vector X is said to be a generalized eigenvector of rank m for A and eigenvalue λ if

(A-λI)m = 0

An ordinary eigenvector of A is also a generalized eigenvector of A because

(A-λI)X = 0
(A-λI)0X = IX = X ≠ 0

The Chain of Vectors Generated from
a Generalized Eigenvector of Rank m

Let Xm be a generalized eigenvector of rank m for matrix A corresponding to the eigenvalue λ. Let Xm-1 be created from Xm as (A-λI)Xm and likewise Xm-2 from Xm-1 and so on down to X1. Thus for j=1, … (m-1)

Xj = (A-λI)Xj+1
and hence
Xj = (A-λI)m-jXm

The sequence {Xm, Xm-1, … X1} is called the chain generated by the generalized eigenvector Xm. Each element Xj of the chain is a generalized eigenvector of A associated with its eigenvalue λ. Furthermore the rank of Xj is j.


By definition

(A-λI)mXm = 0

Now consider (A-λI)jXj and (A-λI)j-1Xj.

(A-λI)jXj = (A-λI)j(A-λI)m-jXm = (A-λI)m Xm = 0
(A-λI)j-1Xj = (A-λI)j-1(A-λI)m-jXm = (A-λI)m-1 Xm0

The vectors in the chain generated by a generalized
eigenvector Xm of a matrix A and its
eigenvalue λ are linearly independent.


Suppose for a set of scalars {cm, cm-1, … c1}

cmXm + cm-1Xm-1 + … + c1X1 =0

For later reference this will be called the dependence condition.

Linear independence requires that cj=0 for all j from m down to 1.

First multiply on the left the above equation by (A-λI)m-1 . Every term for j=m-1 down to 1 becomes the zero vector 0 because

(A-λI)m-1cjXj = cj(A-λI)m-1Xj
which can be rewritten as
cj(A-λI)m-1-j(A-λI)jXj = cj(A-λI)m-1-j0 = cj0 = 0

Therefore the equation involving the cj's reduces to

cm(A-λI)m-1 Xm = 0
but (A-λI)m-1 Xm0 so
cm must be 0

Now multiply the dependence condition by (A-λI)m-2. By the same reasoning as above this leads to the conclusion that cm-1=0.

The procedure may be repeated until it is established that cj=0 for all j from m down to 1. Thus the vectors of the chain are linearly independent.

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