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
but
(A-λI)^m-1 ≠ 0

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

(A-λI)X = 0
but
(A-λI)⁰X = IX = X ≠ 0

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

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

X_j = (A-λI)X_j+1
and hence
X_j = (A-λI)^m-jX_m

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

Proof:

By definition

(A-λI)^mX_m = 0
but
(A-λI)^m-1X_m ≠ 0

Now consider (A-λI)^jX_j and (A-λI)^j-1X_j.

(A-λI)^jX_j = (A-λI)^j(A-λI)^m-jX_m = (A-λI)^m X_m = 0
but
(A-λI)^j-1X_j = (A-λI)^j-1(A-λI)^m-jX_m = (A-λI)^m-1 X_m ≠ 0

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

Proof:

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

c_mX_m + c_m-1X_m-1 + … + c₁X₁ =0

For later reference this will be called the dependence condition.

Linear independence requires that c_j=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-1c_jX_j = c_j(A-λI)^m-1X_j
which can be rewritten as
c_j(A-λI)^m-1-j(A-λI)^jX_j = c_j(A-λI)^m-1-j0 = c_j0 = 0

Therefore the equation involving the c_j's reduces to

c_m(A-λI)^m-1 X_m = 0
but (A-λI)^m-1 X_m≠0 so
c_m 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 c_m-1=0.

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