algebraic and geometric multiplicity do not coincide

Zero is an eigenvalue of

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$\ker A$. Thus, suppose

This implies $b=0$, so

$$\ker A=\mathrm{span}\left(\begin{array}{c}\hfill 1\hfill \\ \hfill 0\hfill \end{array}\right),$$

and the geometric multiplicity of $0$ is $1$.