# 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.
• 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