Mo logo [home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] german flag
Mathematics-Online course: Linear Algebra - Normal Forms - Jordan Normal Form

Generalized Eigenvectors

[previous page] [next page] [table of contents][page overview]
Let $ \lambda$ be an eigenvalue of matrix $ A$ with algebraic multiplicity $ m$. A vector $ v$ with

$\displaystyle (A-\lambda E)^mv=0\,,\quad v\neq 0 $

is called generalised eigenvector for eigenvalue $ \lambda$. All generalised eigenvectors together with the zero vector form a subspace of dimension $ m$ called generalised eigenspace $ H_\lambda$ for eigenvalue $ \lambda$. This subspace is invariant under the linear mapping $ A$.
Since the identity matrix commutes with each matrix we obtain for the image $ Av$ of a vector $ v \in H_\lambda$:
$\displaystyle (A-\lambda E)^mAv$ $\displaystyle =$ $\displaystyle \left(A^{m+1}+\cdots +\lambda^mE^mA\right)v= A\left(A^{m}+\cdots +\lambda^mE^m\right)v$  
  $\displaystyle =$ $\displaystyle A(A-\lambda E)^mv=A\cdot 0=0\,.$  

Consequently, we have $ Av\in H_\lambda$.

In order to find the dimension we bring matrix $ A$ to upper triangle form:

$\displaystyle A \to B = Q^{-1} A Q = \left(\begin{array}{cccccc} \lambda & & *... ...a& * & & * \\ {\qquad} \\ & 0 & & & R & \\ {\qquad} \end{array}\right) \,. $

For $ v \in H_\lambda$ we have

$\displaystyle 0 = Q^{-1} (A-\lambda E)^m Q (Q^{-1} v) = (B-\lambda E)^m (Q^{-1} v)\,, $

thus, the generalised eigenvectors transform according to $ v\to w=Q^{-1}v$. By the form of $ B$ it can easily be seen that for $ i\le m$ the unit vectors $ e_i$ belong to the generalised eingenspace $ Q^{-1}H_\lambda$ of $ B$ but for von $ i>m$ they do not. Consequently we have $ \operatorname{dim}H_\lambda = m$.
(Authors: Burkhardt/Höllig/Hörner)
  automatically generated 4/21/2005