I´m required to prove (or disprove) the following:
If $A \in M_{k+1}(\mathbb{C}) $ is not the null matrix, $k \ge2$, $A^k=A$, and there exists $\lambda \in \sigma(A)$ with geometric multiplicity equal to $2$, $A$ is diagonalizable.
Roughly I would say it is diagonalizable since $x^k-x$ is polynomial which splits into lineal factors and also annihilates $A$, so $x^k-x$ is divisible by the minimal polynomial of $A$, and then the minimal polynomial itself splits into linear factors.
I´m not sure however what the geometric multiplicity of $\lambda$ tells beyond the fact that the characteristic polynomial should have $(x-\lambda)^2$ as a factor.
The geometric multiplicity of $\lambda$ is the dimension of $\ker(A-\lambda)$. The algebraic is the dimension of $\cup_{k\geq 0} \ker (A-\lambda)^k$. In particular, the algebraic dimension is strictly greater than the geometric if there is a vector $v$ for which $$ w=(A-\lambda)v \neq 0 \ \ \ \mbox{and} \ \ \ (A-\lambda)w = 0 $$ This reflects in the minimal polynomial $m$ for $A$ since if $\lambda$ is a simple root of $m$, i.e. $m(x)=q(x)(x-\lambda)$ with $q(\lambda)\neq 0$ then $m(A)=0$ but
$$ m(A) v = q(A) (A-\lambda) v= q(A) w = q(\lambda) w \neq 0$$ which is a contradiction so such a vector $w$ can not exist in that case. If $A$ admits a minimal polynomial with only simple roots then every eigenvector is then a geometric eigenvector. This implies that the matrix has a basis of geometric eigenvectors and then that $A$ is diagonalizable.