If $(A-\lambda I)^r q(A) =0$ then $q(V) \subset \text{ker}(A-\lambda I)^r$ I was reading this answer to a question about the multiplicity of the roots in the minimal polynomial. The hypothesis of the original question is as follows:


  Let $ V\neq \{0 \}$ be a finite-dimensional vector space over a field $F$ and let $A \in \text{End}(V)$


At the beginning of the answer, I was reading the author establishes the following:


  We have $(A-\lambda I)^r q(A) =0$, which is to say that $q(V) \subset \text{ker}(A-\lambda I)^r$


I understand that the first equation is $0$ because by definition the minimal polynomial annihilates the transformation to which it's related to, but I don't see how this implies the second part. It is my understanding that $ \text{ker}(A-\lambda I)^r$ is the set of all generalized eigenvectors associated to the eigenvalue $\lambda$, so is this saying that any vector in $V$ applied to the polynomial $q$ is a generalized eigenvector?

Can someone tell me why this implication holds or tell me what I'm missing? Thank you in advance!

EDIT: The original author of the question edited his answer to have $q(A)V$ instead of $q(A)$, which is what some people already said they suspected in the comments of this question.
Given a linear operator $A : V \to V$ and a polynomial $q(t)$ such that $(A - \lambda I)^r q(A) = 0,$ it follows that $(A - \lambda I)^r q(A)(v) = 0(v) = 0$ for all $v$ in $V,$ hence we have $\operatorname{range}(q(A)) \subseteq \ker(A - \lambda I)^r.$ Considering that $\operatorname{range}(q(A)) = q(A)(V)$ by definition, we conclude that $q(A)(V) \subseteq \ker(A - \lambda I)^r.$
The kernel of an operator $T\in \Bbb L(V$) is defined by $\{v\in V \mid T(v)=0\}$. Hence $(A-\lambda I)^r q(A) =0$ is exactly the same as saying every  $q(A)(v) \subset \text{ker}(A-\lambda I)^r\; \forall v\in V$ 

(I believe the V was supposed to be A. I've learned this part in Linear Algebra but we never used q(V) with V a vector space.)