# Prove that $q \leq \dim(E_\lambda)$ given that $J$ has $q$ Jordan blocks associated with $\lambda$

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 $\lambda$. – levap Dec 6 '12 at 16:36

## 2 Answers

Every Jordan block for $~\lambda$ has an eigenspace for$~\lambda$ of dimension$~1$ spanned by the first vector of$~B$ associated to this block. Since the $q$ such Jordan blocks form a direct sum, the sum of their eigenspaces for$~\lambda$ is also direct, and it is contained in the eigenspace$~E_\lambda$; this gives you $q\leq\dim(E_\lambda)$. In fact $E_\lambda$ is the sum of the eigenspaces of those Jordan blocks, so you even have $q=\dim(E_\lambda)$.

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From the Jordan form of a linear map $T$ you can read the dimensions of the regular and generalized eigenspaces. Each Jordan block for the eigenvalue $\lambda$ looks like $$\left( \begin{array}{cccc} \lambda & 1 & \ldots & 0 \\ 0 & \lambda & 1 & 0 \\ 0 & 0 & \lambda & 1 \\ 0 & 0 & 0 & \lambda \end{array} \right).$$

If the block is $k \times k$, it comes with exactly one (linearly independent) eigenvector $e_1$ and $k - 1$ generalized eigenvectors $e_2, ..., e_k$.

If in the Jordan block decomposition of $J$, there are exactly $q$ blocks of this form, each of size $k_i \times k_i$, then there are exactly $q$ (linearly independent) eigenvectors with corresponding eigenvalue $\lambda$, so you have in fact equality.

Try also to see how to read the dimension of $\ker(T - \lambda I)^2$ only from the number of Jordan blocks and their size.

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Ok, so if I try to read the $dim(ker(T-\lambda I)^2)$, then that should be 1 as well. Correct? – tk2 Dec 6 '12 at 17:02
Yeah. So if you have a matrix $A$ with two blocks of size $3 \times 3$ and one block of size $1 \times 1$ associated to $\lambda$, you have $\dim(\ker(T - \lambda I)) = 3$, $\dim(\ker(T - \lambda I)^2) = 2$, and $\dim(\ker(T - \lambda I)^3) = 2$, which is consistent with the fact that the matrix should be $7 \times 7$. – levap Dec 6 '12 at 17:07
@levap Those last two dimension are wrong, since the kernels contain the previous ones; the numbers are $3$, $3+2=5$, and $3+2+2=7$ respectively. – Marc van Leeuwen Nov 27 '15 at 7:39
@Marc van Leeuwen I can't edit my comment anymore but you are of course correct. Thanks! – levap Nov 27 '15 at 13:34