# difference between eigenspace and generalized eigenspace

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.

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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. 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.
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.