What is the relation between rank of a matrix, its eigenvalues and eigenvectors I am quite confused about this. I know that zero eigenvalue means that null space has non zero dimension. And that the rank of matrix is not the whole space. But is the number of distinct eigenvalues ( thus independent eigenvectos ) is the rank of matrix?
 A: Well, if $A$ is an $n \times n$ matrix, the rank of $A$ plus the nullity of $A$ is equal to $n$; that's the rank-nullity theorem. The nullity is the dimension of the kernel of the matrix, which is all vectors $v$ of the form:
$$Av = 0 = 0v.$$
The kernel of $A$ is precisely the eigenspace corresponding to eigenvalue $0$. So, to sum up, the rank is $n$ minus the dimension of the eigenspace corresponding to $0$. If $0$ is not an eigenvalue, then the kernel is trivial, and so the matrix has full rank $n$. The rank depends on no other eigenvalues.
A: My comment is 7 years late but I hope someone might find some useful information.
First, the number of linearly independent eigenvectors of a rank $k$ matrix can be greater than $k$.
For example
\begin{align}
A &= \left[ \begin{matrix}
1 & 2 \\
2 & 4
\end{matrix} \right] \\
rk(A) &= 1 \\
\end{align}
$A$ has the following eigenvalues and eigenvectors
$\lambda_1 = 5, \mathbf{v}_1 = [1 \ \ 2]^\top$, $\lambda_2 = 0, \mathbf{v}_2 = [-2 \ \ 1]^\top$.
So $A$ has 1 linearly independent column but 2 linearly independent eigenvectors.
The column space of $A$ has 1 dimension. The eigenspace of $A$ has 2 dimensions.
There are also cases where the number of linearly independent eigenvectors is smaller than the rank of $A$. For example
\begin{align}
A &= \left[ \begin{matrix}
0 & 1 \\
-1 & 0
\end{matrix} \right] \\
rk(A) &= 2
\end{align}
$A$ has no real valued eigenvalues and no real valued eigenvectors. But $A$ has two complex valued eigenvalues $\lambda_1 = i,\ \lambda_2 = -i$ and two complex valued eigenvectors.
Another remark is that a eigenvalue can correspond to multiple linearly independent eigenvectors. An example is the Identity matrix. $I_n$ has only 1 eigenvalue $\lambda = 1$ but $n$ linearly independent eigenvectors.
So to answer your question, I think there is no trivial relationship between the rank and the dimension of the eigenspace.
