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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am studying Generalized Eigenvectors. It seems that we can define them as $\mathbf{p}_i$ in this equation:

$$ (\mathbf{A}-\lambda\mathbf{I})^{k}\mathbf{p}_i = \mathbf{0} $$

in which $k$ is the algebraic multiplicity of $\lambda$ in $ |\mathbf{A}-\lambda\mathbf{I}|=0$. Also it can be defined as:

$$ (\mathbf{A}-\lambda\mathbf{I})\mathbf{p}_i = \mathbf{p}_{i-1},~~i=1\ldots,~~\mathbf{p}_{0}=\mathbf{0} $$

or at least this is what I have learned (if sth is wrong, please let me know).

Are these definitions equivalent? I mean $ (\mathbf{A}-\lambda\mathbf{I})^{k}\mathbf{p}_i = \mathbf{0} $ if and only if $ (\mathbf{A}-\lambda\mathbf{I})\mathbf{p}_i = \mathbf{p}_{i-1} $?

I can prove the first one, if the second one is true (by multiplying both sides in $ (\mathbf{A}-\lambda\mathbf{I})$ k times), but how can I prove the second one, given the first one?


share|cite|improve this question

The more general (common?) definition is the first one you gave. A non-zero vector $\mathbf{v}$ is a generalized eigenvector if and only if $(A-\lambda I)^k \mathbf{v} = \mathbf{0}$ for some $k\in \mathbb{N}^+$.

The second definition is more commonly used to construct the chains of generalized vectors for the Jordan normal form.

Suppose that $\mathbf{v}$ is a generalized eigenvector. Then there exists some $k$ such that $$(A-\lambda I)^k \mathbf{v} = \mathbf{0}$$ Further suppose that $k$ is the smallest such integer. Then we define $$\mathbf{v}_i = (A-\lambda I)^{k-i}\mathbf{v}$$ and we call this a chain of generalized eigenvectors of length $k$ $$\mathcal{C} = \{\mathbf{v}_1,\ \cdots,\ \mathbf{v}_{k}\}$$ You can easily verify that $\mathbf{v}_k = \mathbf{v}$ and that $\mathbf{v_1}$ is in fact an eigenvector of $\lambda$. For each generalized eigenvector, there will be associated such a chain of generalized eigenvectors, the length of the chain corresponding to the index of the vector. Conversely, if you build such a chain from an eigenvector then each member of the chain will be a generalized eigenvector.

These chains are what determines the Jordan block structure. Each eigenvector will have a chain associated with it and if the eigenvectors leading the chains are linearly independent then so are the chains that they generate. The number of linearly independent chains determine the number of Jordan blocks and the size of the chains determine the size of the corresponding blocks.

share|cite|improve this answer
Thanks a lot.2 questions: Am I right about k (is it algebraic multiplicity of λ or just some + integer?) and are these two definitions equivalent? – Ramin Dec 8 '12 at 9:06
The two definitions are equivalent. The construction I gave above shows that the first definition implies the second and you said that you managed to prove that the second implies the first. As for $k$, the algebraic multiplicity is guaranteed to work but $k$ itself depends on the vector $\mathbf{v}$. Different generalized eigenvectors will be annihilated at different depths. The smallest exponent which is guaranteed to eliminated all generalized eigenvectors is actually not the algebraic multiplicity but rather the exponent of $(x-\lambda)$ in the minimal polynomial. – EuYu Dec 8 '12 at 9:16
@BarbaraOsofsky Yes, my apologies. I should've mentioned that they are non-zero. – EuYu Feb 1 '13 at 7:11

While it is true that generalized eigenvectors appear in many calculations with an index $i$ such an index should not appear in their definition.

From a taxonomical (not a didactical) standpoint the proper order of definitions would be as follows:

Given a vector space $V$ over a field $K$ and a linear transformation $A:\ V\to V$, a nonzero vector ${\bf p}\in V$ is called a generalized eigenvector of $A$ if there is a $\lambda\in K$ and a $k\in{\mathbb N}_{\geq1}$ such that $$(A-\lambda I)^k\>{\bf p}={\bf 0}\ .$$ When $k=1$ the vector ${\bf p}$ is simply called an eigenvector.

share|cite|improve this answer

The definitions are not equivalent. Let $\mathbf{A}$ be the $n$ by n identity matrix. Then $\mathbf {A - I = 0}$. The problem is that there are several things involved here, the characteristic polynomial, the minimal polynomial, and the elementary divisors. The generalized eigenspace for eigenvalue $\lambda$ has, as you say, dimension $k$ where $k$ is the largest power of $(X-\lambda)$ that divides the characteristic polynomial. The elementary divisors are a special set of powers of $(X-\lambda)$ whose exponents add up to $k$.

Your second definition works for each elementary divisor $(X-\lambda )^d$: you can find a vector $\mathbf {p}$ which is a generalized eigenvector not appearing in any other space corresponding to an elementary divisor, such that a basis for the corresponding generalized eigenspace is $$\mathbf p,\ (A-\lambda I)(\mathbf p), \ \cdots ,\ (A-\lambda I)^{d-1}(\mathbf p).$$ When you get to generalized eigenvectors you will begin working with the second "definition" but until your instructor introduces the topic of elementary divisors and Jordan blocks, concentrate on when all the eigenvectors span the vector space.

share|cite|improve this answer
I don't see why $A=I$ is a counterexample. If $A=I$ then according to both definitions, everything is a generalized vector. – Ted Feb 1 '13 at 6:55
Not quite. The zero vector is is never an eigenvector of any kind by definition. If you pick a nonzero vector and multiply it by $\mathbf I-I$ you will get a zero vector. The elementary divisors are all $X-i$ here, and a linearly independent set of eigenvectors consists of the columns of the identity matrix. – Barbara Osofsky Feb 1 '13 at 7:03

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.