Re: Matrix Diagonalization

On Sun, 15 Mar 2009 12:01:15 -0700, rip pelletier wrote:

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

Thanks interesting stuff, rip. I took a look at the wiki, and it's *very*
interesting and visually pimped-out. I find myself pining away for the
text I had for the Mathematical Methods class that seniors in physics take.
I was a little shaky to equate the appropriate determinant to zero.

For those who almost remember this stuff, with

( 1 0 )
( 1 1 )
, the characteristic equation is
(1-£)(1-£)- (1*0) = 0
To ascertain the roots, you can wave your hands to get rid of the term
that's multiplied by zero, and then you have a fully factored expression on
the LHS.

I think the question at hand is what to do with eigenvalue algorithms,
when, from a theory of equations point of view, there is a root with
multiplicity.

If Lapack does a good job giving you *distinct* roots, then I think we need
to see how many this root satisfies the characteristic equation.

What I think we should look at is a three by three with exactly two equal,
real roots and see how lapack does and what we have to do to make it
robust.
--
larry gates

Perl will always provide the null.
-- Larry Wall in <199801151818.KAA14538@xxxxxxxx>
.

Relevant Pages

  • 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: Matrix Diagonalization
    ... The all zeroes is not invalid, it is the 'generilized' eigenvector. ... a generalized eigenvector is something else. ... The diagonalization of A is ... form (JCF). ...
    (comp.lang.fortran)
  • Re: Cononical Form
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  • Re: roots of determinant of toeplitz-like matrix
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