Re: Matrix Diagonalization

In article <8208a$49bb86df$d55d2be5$2068@xxxxxxxxxxxxxx>,
Jan Gerrit Kootstra <jan.gerrit@xxxxxxxxxxxxxxx> wrote:

The all zeroes is not invalid, it is the 'generilized' eigenvector. So
it is not what one should expect from a classical point of view.

Actually, the zero vector is neither an eigenvector nor a generalized
eigenvector. After all, if it were then every complex number z would be
an eigenvalue associated with it. Okay, maybe someone has some use for
_everything_ being an eigenvalue; but as soon as we think that we are
only interested in the usual eigenvalues, we exclude the zero vector as
an eigenvector of any kind.

More to the point, a generalized eigenvector is something else.

As an eigenvector v of a matrix A satisfies the equation (A - e I)v = 0
where e is the eigenvalue associated with v, a generalized eigenvector v
satisfies the equation (A - e I)^2 v = 0 (more generally, the equation
(A - e I)^k v = 0).

That takes care of computing them, in principle. (Practice is certainly
another issue.) But how do we understand them? Well, the question I can
answer is: how do _I_ understand them?

An nxn matrix A can be diagonalized if and only if we can find n
linearly independent eigenvectors. From them we construct a matrix P
whose columns are the eigenvectors. The diagonalization of A is
accomplished by the similarity transformation

P^-1 A P,

which produces a diagonal matrix.

What if the matrix A cannot be diagonalized? Well, we "know" that there
is a similarity transformation Q which will bring it to Jordan canonical
form (JCF). By hypothesis, we don't have enough eigenvectors to build Q;
but there must exist other vectors which can be used as additional
columns, to fill it out.

Those are the generalized eigenvectors.

(My undergraduate class focused on the existence of the JCF, and never
mentioned the similarity transformation Q that brings it about. But for
me, it's Q that makes sense of the whole thing. In fact, I was long out
of grad school before I ever heard of generalized eigenvectors. My
advanced classes needed only the existence of the JCF, and didn't care
how it was gotten.)

For the example given (2x2, lower triangular, all 1s), the eigenvector
is any multiple of (0,1); the generalized eigenvector is _any_ nonzero
2D vector. In particular, (1,0) is a nice choice (but anything not
proportional to the eigenvector (0,1) suffices). For this example the
result is almost trivial. It is not generally the case that the entire
space is the generalized eigenspace.

If (A - e I)v = 0 doesn't have enough linearly independent solutions v,
we look at (A - e I)^2 v = 0 for more. Keep going until we have enough.

By the way, in this example (A - e I)^2 is the zero matrix, so higher
powers k > 2 are irrelevant.

A pleasant and applied reference is

Perko, Lawrence. <b>Differential Equations and Dynamical Systems.</b>
Springer, 2001 (3rd ed). ISBN 0 387 95116 4

I happen to know of one blog post which discusses the geometry of
generalized eigenvectors, but I have not done a search.

http://unapologetic.wordpress.com/2009/02/19/the-multiplicity-of-an-eigen
value/

vale,
rip

--
NB eddress is r i p 1 AT c o m c a s t DOT n e t
.

Relevant Pages

  • Re: Matrix Diagonalization
    ... The all zeroes is not invalid, it is the 'generilized' eigenvector. ... a generalized eigenvector is something else. ... The diagonalization of A is ... If Lapack does a good job giving you *distinct* roots, ...
    (comp.lang.fortran)
  • Re: Cononical Form
    ... >> books use the notation trto denote a column vector!) ... > Can you see how, if you know the matrix, the eigenvector, and the ... >>> Your notation makes no sense. ... >>> a generalized eigenvector. ...
    (sci.math)
  • Re: Cononical Form
    ... this would be the smallest such power such that we obtain ... so you can pick anything that isn't an eigenvector. ... >>> Please don't toppost. ... >>> for a vector to be a generalized eigenvector? ...
    (sci.math)
  • Re: Help with eigenvalue decomposition
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