What is the relation between the eigenspace of a matrix and its column space? I was wondering if there is any relation between the space span by all eigenvector of a matrix $A$ and its column space.
Also, is there any condition on $A$ that ensures that these two spaces are the same?
 A: 
is there any condition on A so that these two spaces are the same

Yes.  Over a field $F$ of scalars, the condition is that


*

*all eigenvalues of $A$ belong to $F$ (all irreducible factors of the characteristic polynomial are linear)

*Every element of the $0$-eigenspace, is in the image of $A$ (there is a direct sum decomposition of the generalized $0$-eigenspace into $2 \times 2$ examples like the one in Gerry Myerson's answer) 

*every generalized eigenvector with nonzero eigenvalue is an eigenvector (if $\lambda \neq 0$ is a root of multiplicity $m$ in the characteristic polynomial, the $\lambda$ eigenspace of $A$ has dimension $m$)
A: To add a bit to Gerry's answer... 
If you look at a matrix as a linear operator, $T(v)=Av$ then the column space is just the range of that linear operator. Eigenvectors for non-zero eigenvalues will be members of the range (if $Av=\lambda v$, then $A(\lambda^{-1}v) = v$). 
So the span of the eigenvectors with non-zero eigenvalues, is contained in the column space.
(...and the span of the eignevectors with eigenvalue zero is the null space.)
A: [Comment elevated to answer]
In general, there is no relation between the two spaces. 
For example, if $$A=\pmatrix{1&1\cr0&1}$$ then the column space is all of ${\bf R}^2$ while the eigenvectors span a 1-dimensional subspace. 
If $$A=\pmatrix{0&0\cr0&0\cr}$$ then the eigenvectors span all of ${\bf R}^2$ while the column space is just the zero vector. 
If $A$ is invertible and diagonalizable, then the two spaces will both be all of ${\bf R}^n$. I expect there are other examples of equality. 
EDIT: Here's another example of equality: $$A=\pmatrix{2&-4\cr1&-2\cr}$$ The vector $(2,1)$ generates the column space, and also generates all the eigenvectors. 
A: Let me point out the situation for the special case where A is symmetric positive definite:


*

*If a matrix A $\in \mathbb{R}^{n \times n}$ is symmetric positive definite, then it diagonalizable, as the Schur transform gives a diagonal matrix. 

*Thus, the algebraic multiplicity equals the geometric multiplicity for every eigenvalue. 

*As the eigenvectors of Hermitian matrices are orthogonal to each other, we know that the dimension of the span of the eigenvectors is equal to the sum of the geometric multiplicities over the different eigenvalues which is $n$.


Thus, the eigenvectors span $\mathbb{R}^n$ in this special case.
