Question about multiplicity of eigenvalue = 0 for singular matrix

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?

• That matrix has characteristic polynomial $(3-\lambda)^2 - 1$, whose roots are $2$ and $4$. – Kaj Hansen Jul 26 '19 at 20:29
• @KajHansen I think the characteristic polynomial is just $(3-\lambda)^2$? At least that's what I get from this – Yandle Jul 27 '19 at 23:02

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.

• My question arises from this video where the professor had a rank 1 3x3 matrix and immediately wrote down eigenvalue = 0 twice after stating that the nullity = 2. I'm guessing for this example he knows the third eigenvalue one is not zero because the trace is 1? – Yandle Jul 28 '19 at 21:40
• If rank is $1$ at least two eigenvalues are zeros (algebraic multiplicity but they could be three zeros), yes it means $\ker(matrix)$ has dimension two (geometric multiplicity) the last eigenvalue is concluded as you said. – Toni Mhax Jul 28 '19 at 22:06

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