For an $n\times n$ matrix $A$ of rank $r$, for $\lambda = 0$, I think the dimension of the eigenspace (equal to the null space of $A$) is always $n-r$. Is it possible to show whether the multiplicity of $\lambda = 0$ can exceed $n-r$?
If the multiplicity of $\lambda = 0$ always equal the dimension of its eigenspace ($n-r$), why is this true yet for a matrix like $\begin{bmatrix}3&1\\0&3\end{bmatrix}$, we have $\lambda = 3$ with a multiplicity of 2 but only one eigenvector?
On
You may be under the impression that R^n is spanned by the eigenvectors of a matrix this is actually not true however see https://en.wikipedia.org/wiki/Jordan_normal_form
It is typically true however one can think of large blocks in the JOrdan normal form as a "degeneracy" so to speak.
For a simple counter example to your claim consider the right shift operator where the right most element is deleted one can easily see any eigenvalue must be 0 but the dimension of the nullspace is merely one
you may also want to look up idempotent and nilpotent operators
The multiplicity of an eigenvalue known as algebraic multiplicity is $\ge $ than the geometric multiplicity (geometric multiplicity is $n-r$ for your exemple of $\lambda=0$). A classic fact.