Zero is an eigenvalue of $$ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} $$ with algebraic multiplicity $2$ and geometric multiplicity $1$ .

Indeed, as $$ \det (A-\lambda I) = \lambda^2 $$ it follows that $0\,\!$ is an eigenvalue of $A$ with algebraic multiplicity $2$ . To find the geometric multiplicity of $A$ we need to calculate $\operatorname{ker} A$ . Thus, suppose $$ \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}. $$ This implies $b=0$ , so $$ \ker A = \operatorname{span} \begin{pmatrix} 1 \\ 0 \end{pmatrix}, $$ and the geometric multiplicity of $0\,\!$ is $1$ .