San José State University


appletmagic.com Thayer Watkins Silicon Valley & Tornado Alley U.S.A. 

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
An ordinary eigenvector of A is also a generalized eigenvector of A because
Let X_{m} be a generalized eigenvector of rank m for matrix A corresponding to the eigenvalue λ. Let X_{m1} be created from X_{m} as (AλI)X_{m} and likewise X_{m2} from X_{m1} and so on down to X_{1}. Thus for j=1, … (m1)
The sequence {X_{m}, X_{m1}, … X_{1}} 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
Now consider (AλI)^{j}X_{j} and (AλI)^{j1}X_{j}.
Proof:
Suppose for a set of scalars {c_{m}, c_{m1}, … c_{1}}
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 (AλI)^{m1} . Every term for j=m1 down to 1 becomes the zero vector 0 because
Therefore the equation involving the c_{j}'s reduces to
Now multiply the dependence condition by (AλI)^{m2}. By the same reasoning as above this leads to the conclusion that c_{m1}=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.
HOME PAGE OF appletmagic 
