generalized eigenvector

Let V be a vector spaceMathworldPlanetmath over a field k and T a linear transformation on V (a linear operator). A non-zero vector vV is said to be a generalized eigenvectorMathworldPlanetmath of T (corresponding to λ) if there is a λk and a positive integer m such that


where I is the identity operatorMathworldPlanetmath.

In the equation above, it is easy to see that λ is an eigenvalueMathworldPlanetmathPlanetmathPlanetmathPlanetmath of T. Suppose that m is the least such integer satisfying the above equation. If m=1, then λ is an eigenvalue of T. If m>1, let w=(T-λI)m-1(v). Then w0 (since v0) and (T-λI)(w)=0, so λ is again an eigenvalue of T.

Let v be a generalized eigenvector of T corresponding to the eigenvalue λ. We can form a sequence


The set Cλ(v) of all non-zero terms in the sequence is called a cycle of generalized eigenvectors of T corresponding to λ. The cardinality m of Cλ(v) is its . For any Cλ(v), write vλ=(T-λI)m-1(v).

Below are some properties of Cλ(v):

  • vλ is the only eigenvectorMathworldPlanetmathPlanetmathPlanetmath of λ in Cλ(v), for otherwise vλ=0.

  • Cλ(v) is linearly independentMathworldPlanetmath.


    Let vi=(T-λI)i-1(v), where i=1,,m. Let 0=i=1mrivi with rik. Induct on i. If i=1, then v1=v0, so r1=0 and {v1} is linearly independent. Suppose the property is true when i=m-1. Apply T-λI to the equation, and we have 0=i=1mri(T-λI)(vi)=i=1m-1rivi+1. Then r1==rm-1=0 by inductionMathworldPlanetmath. So 0=rmvm=rmvλ and thus rm=0 since vλ is an eigenvector and is non-zero. ∎

  • More generally, it can be shown that Cλ(v1)Cλ(vk) is linearly independent whenever {v1λ,,vkλ} is.

  • Let E=span(Cλ(v)). Then E is a (m+1)-dimensional subspacePlanetmathPlanetmath of the generalized eigenspaceMathworldPlanetmath of T corresponding to λ. Furthermore, let T|E be the restrictionPlanetmathPlanetmath of T to E, then [T|E]Cλ(v) is a Jordan blockMathworldPlanetmath, when Cλ(v) is ordered (as an ordered basis) by setting

    (T-λI)i(v)<(T-λI)j(v)   whenever   i>j.

    Indeed, for if we let wi=(T-λI)m+1-i(v) for i=1,m+1, then

    T(wi)=(T-λI+λI)(T-λI)m+1-i(v) = {λwi if i=1,wi-1+λwi otherwise.

    so that [T|E]Cλ(v) is the (m+1)×(m+1) matrix given by

  • A cycle of generalized eigenvectors is called maximal if v(T-λI)(V). If V is finite dimensional, any cycle of generalized eigenvectors Cλ(v) can always be extended to a maximal cycle of generalized eigenvectors Cλ(w), meaning that Cλ(v)Cλ(w).

  • In particular, any eigenvector v of T can be extended to a maximal cycle of generalized eigenvectors. Any two maximal cycles of generalized eigenvectors extending v span the same subspace of V.


  • 1 Friedberg, Insell, Spence. Linear Algebra. Prentice-Hall Inc., 1997.
Title generalized eigenvector
Canonical name GeneralizedEigenvector
Date of creation 2013-03-22 17:23:13
Last modified on 2013-03-22 17:23:13
Owner CWoo (3771)
Last modified by CWoo (3771)
Numerical id 13
Author CWoo (3771)
Entry type Definition
Classification msc 65F15
Classification msc 65-00
Classification msc 15A18
Classification msc 15-00
Related topic GeneralizedEigenspace
Defines cycle of generalized eigenvectors