S.O.S. Mathematics CyberBoard

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.

_________________
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.

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.

Time to go re-learn Jordan Canonical form...

Thanks

_________________
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.