During exercise classes for Jordan normal form, we said that if $\dim\ker((A - \lambda_1 I)^{a_1}(A - \lambda_2 I)^{a_2}) = k$, then $\dim\ker(A - \lambda_1 I)^{a_1} + \dim\ker(A - \lambda_2 I)^{a_2}=k$.
(Where $A \in \mathbb{C}^{n \times n}$, $\lambda_1, \lambda_2$ are different eigenvalues of $A$ and $a_1, a_2 \in \mathbb{N}$)
I don't understand why that is true, I know that for every two linear maps $C,D$ it holds that $\dim\ker(AB) \leq \dim\ker(A) + \dim\ker(B)$, but I don't know why the equality holds.