The assumption implies $\mathrm{ran}(A)=\mathrm{ran}(A^k)$, hence also $\mathrm{ker}(A)=\mathrm{ker}(A^k)$, for all $k\in\mathbb{N}$. The algebraic multiplicity of an eigenvalue $\lambda$ of the matrix $A$ is $\max _ {k\in\mathbb{N}} \mathrm{dim} \ker (A-\lambda)^k $ (think e.g. to the Jordan form of $A$). So in your case the algebraic and geometric multiplicity of the eigenvalue $0$ coincide.

The assumption implies $\mathrm{ran}(A)=\mathrm{ran}(A^k)$, hence also $\mathrm{ker}(A)=\mathrm{ker}(A^k)$, for all $k\in\mathbb{N}$. The algebraic multiplicity of an eigenvalue $\lambda$ of the matrix $A$ is $\max _ {k\in\mathbb{N}} \mathrm{dim} \ker (A-\lambda)^k $ (think e.g. to the Jordan form of $A$). So in your assumption the algebraic and geometric multiplicity of the eigenvalue $0$ coincide (and conversely).