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The Generalized Eigenvectors of a Matrix and their Linear Indepedence |
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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 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)
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.
Proof:
By definition
Now consider (A-λI)jXj and (A-λI)j-1Xj.
Proof:
Suppose for a set of scalars {cm, cm-1, … c1}
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
Therefore the equation involving the cj's reduces to
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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