Take the 2-minute tour ×
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.

What are the differences between eigenspace and generalized eigenspace? Why do we need generalized eigenspace? Can an arbitrary matrix (not necessarily over $\mathbb{C}$) have a Jordan form? Thank you very much.

share|improve this question
1  
1) Do you know the definitions? The definitions are different, and it is not hard to find an example of a generalized eigenspace which is not an eigenspace by writing down any nontrivial Jordan block. 2) Because eigenspaces aren't big enough in general and generalized eigenspaces are the appropriate substitute. 3) Yes, by passing to the algebraic closure, or by changing somewhat the definition of Jordan normal form. See en.wikipedia.org/wiki/… . –  Qiaochu Yuan May 31 '11 at 14:52

1 Answer 1

up vote 4 down vote accepted
  1. The eigenspace of (a square matrix) $A$ corresponding to $\lambda$ is the collection of all vectors $\mathbf{x}$ that satisfy $A\mathbf{x}=\lambda\mathbf{x}$, or equivalently, $(A-\lambda I)\mathbf{x}=\mathbf{0}$. The generalized eigenspace of $A$ corresponding to $\lambda$ is the collection of all vectors $\mathbf{x}$ for which there exists a positive integer $k$ for which $(A-\lambda I)^k\mathbf{x}=\mathbf{0}$. The former is contained in the latter, but need not be equal. For example, with $$A = \left(\begin{array}{cc}1&1\\0&1\end{array}\right),$$ and $\lambda=1$, you can check easily that the eigenspace consists only of the vectors of the form $(x,0)$ for some arbitrary $x$; whereas the generalized eigenspace is the larger collection of all vectors $(x,y)$, with $x$ and $y$ both arbitrary.

  2. We "need" generalized eigenspaces because the eigenspaces in general do not suffice to describe the entire space (the generalized eigenspaces may not suffice either).

  3. A matrix $A$ with coefficients in a field $F$ has a Jordan canonical form (over $F$) if and only if the characteristic polynomial of $A$ splits over $F$. In particular, all matrices over $F$ have a Jordan canonical form over $F$ if and only if $F$ is algebraically closed. But it's certainly possible to have specific matrices, say over $\mathbb{R}$, that have a Jordan canonical form.

share|improve this answer

Your Answer

 
discard

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.