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 $A$?


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 $2$.


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