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 Post subject: Algebraic vs. Geometric Multiplicity of an eigenvalue...Posted: Tue, 18 Nov 2008 00:35:39 UTC
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Does anyone know where I can find a proof that the geometric multiplicity of an eigenvalue is always less or equal to the algebraic multiplicity of an eigenvalue?

That is if there is a linearly independent set of k eigenvectors, all associated to the eigenvalue e, then (x-e)^k divides the charecteristic polynomial.

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 Post subject: Posted: Tue, 18 Nov 2008 01:55:26 UTC
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Joined: Wed, 1 Oct 2003 04:45:43 UTC
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I don't have any references for you, but I can tell you where to start.

To summarize the situation, we have a vector space V, a linear map T:V->V, and T has precisely k independent eigenvectors associated with the eigenvalue e.

In a nutshell, each generalized eigenspace of T represents a block in the Jordan canonical form of T. Because we have a regular eigenspace of dimension k associated with e, one of the blocks in the Jordan canonical form will be a diagonal matrix of size k with e's along the diagonal. This in turn means that (x-e)^k is a factor of the characteristic polynomial of T.

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 Post subject: Posted: Tue, 18 Nov 2008 02:44:59 UTC
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Joined: Wed, 6 Apr 2005 01:33:29 UTC
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Location: Virginia Beach, VA
Time to go re-learn Jordan Canonical form...

Thanks

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It's quiet now; and as I think my thoughts alone,
I try to keep my head straight, but I think I'm too far gone.
For in this silence, the truth rings even louder.
A constant grinding, begging recognition of its power.

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