Take for a simple example, $A = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$
The geometric multiplicity of an eigenvalue is the number of linearly independent eigenvectors associated with that eigenvalue.
In this case, the eigenvectors corresponding to the zero eigenvalue satisfies,
$Av = 0$
Therefore we may arbitarily pick our eigenvectors. There will be two eigenvectors associated with $0$ eigenvalue. However, we can pick them so they are linearly independent, or not.
In this case, what is the geometric multiplicity of $\lambda = 0$ corresponding to $A$?
Best Answer
The definition of geometric multiplicity of an eigenvalue should be the dimension of eigenspace of that eigenvalue.
In your case, the whole space is an eigenspace corresponding to the eigenvalue $0$. So the geometric multiplicity of the eigenvalue $0$ is $2$.