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Let $T: V \rightarrow V$ be linear, $V$ is a finite dimensional vector space, and the characteristic polynomial of $T$ splits. Also let $\lambda$ be an eigenvalue of $T$ and $B$ be a Jordan Canonical basis for $V$ with respect to $T$. Suppose that $J=[T]_B$ has $q$ Jordan blocks associated with $\lambda$.

Prove that $q \leq \dim(E_\lambda)$.

I'm having trouble even starting off this proof. I know that $K_\lambda$, the generalized eigenspace corresponding to $\lambda$, has an ordered basis of a union of disjoint cycles of generalized eigenvectors. But I'm not sure how this even relates to $E_\lambda$, since for the $K_\lambda$ and $E_\lambda$ to be equal it must be diagonalizable. Or is this something to do with the initial vectors of the generalized eigenvectors of $T$ corresponding to $\lambda$ and the fact that the union of those generalized eigenvectors is disjoint.

Thanks a lot in advance. I really appreciate any help on this particular problem.

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    Maybe you mean $q$ Jordan blocks that are associated with the eigenvalue $\lambda$? Also, it is better to mention explicitly in the beginning that you use $E_\lambda$ for the eigenspace of the eigenvalue $\lambda$.2012-12-06

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