Generalized eigenvectors/eigenvalues

1. Dec 12, 2011

AxiomOfChoice

Mathematica has this command "Eigensystem[{m,a}]", which (to quote their documentation) "gives the generalized eigenvalues and eigenvectors of m with respect to a." I have never encountered this concept before, ever - that there can be eigenvectors of matrices with respect to other matrices. All I have ever come across is that $\lambda$ is a generalized eigenvalue of $A$ with generalized eigenvector $\vec x$ if there exists some $p \in \mathbb N$ such that $(A-\lambda I)^p\vec x = 0$.

Can someone please explain what it *means* to be a "generalized eigenvalue or eigenvector" of m with respect to a? Maybe it is related to the concept I mentioned above, but if so, I don't see it.

2. Dec 13, 2011

micromass

Staff Emeritus
Maybe it is answered in their documentation??

3. Dec 13, 2011

AxiomOfChoice

The only thing they provide is some dodgy example: If you let

$$A = \begin{pmatrix} 1. & 2. \\ 3. & 4. \end{pmatrix}, \quad B = \begin{pmatrix} 1. & 4. \\ 9. & 16. \end{pmatrix},$$

(and they *HAVE* to be decimal entries; if you just put 1 (instead of 1.), you get an error, which just adds to the mystery), then "Eigensystem[{A,B}]" returns the following two complex eigenvalues and two vectors

\begin{aligned} &\{0.25 + 0.193649 I, 0.25 - 0.193649 I\}, \\ &[-0.848472 + 0.0858378 I, 0.498043 + 0.157099 I],\\ &[-0.848472 - 0.0858378 I, 0.498043 - 0.157099 I] \end{aligned}

...huh?

Last edited by a moderator: Dec 13, 2011
4. Dec 13, 2011

Mute

If I recall correctly a generalized eigensystem is one for which

$$Ax = \lambda Bx$$

for given matrices A and B.

EDIT: See here.

5. Dec 13, 2011

D H

Staff Emeritus
6. Dec 13, 2011

AlephZero

A practical application of all this is the vibration of a flexible structure, where A and B represent the stiffness and mass properties. If the matrices are large and sparse (and often also Hermitian and positive semi-definite), solving the generalized eigenproblem is much more efficient than solving the equivalent problem $B^{-1}Ax = \lambda x$, (assuming B is invertible) because $B^{-1}A$ looks like an arbitrary full non-symmetric matrix with no obvious "special properties" to leading to a more efficient solution.

FWIW there are solution procedures that represent $\lambda$ as a ratio of two numbers, with conventions to represent the "indeterminate" or "infinte" eigenvalues and their corresponding vectors. The vectors are well defined and meaningful as the basis vectors of subspaces, even if the corresponding eigenvalues are not so well defined.