Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question
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 2

Every Jordan block has at least one associated basis vector in $B$ (which is not associated to any other Jordan block), and $\dim(E_\lambda)$ is just the number of elements in $B$. So that gives you $\dim(E_\lambda)=\#B\geq q\times1=q$. That's all there is to it.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.