What is the relation between the eigenspace of a matrix and its column space?

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.)

[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.

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 $m$)