Kernel and geometric multiplicity relation

Question

Say I have a square matrix $A$ with one eigenvalue $\lambda_1$. The minimal polynomial is $(\lambda-\lambda_1)^k$ and $\dim(\ker(A))=\alpha$. What can I know about the geometric multiplicity of $\lambda_1$?

Answer 1

If $\lambda_1=0$, then $\dim(\mathrm{ker}(A))=\alpha$ implies that the geometric multiplicity is $\alpha$.

If $\lambda_1\neq 0$, then $A$ is non-singular (for otherwise $0$ would be an eigenvalue of $A$), hence $\alpha=0$. In this case you need to determine $\dim(\mathrm{ker}(A-\lambda_1I))$ in order to find the geometric multiplicity of $\lambda_1$.