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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.

Robert Lee
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2 Answers2

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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 $q(A)(V) \subseteq \ker(A - \lambda I)^r.$

Dylan C. Beck
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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.)

Divide1918
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  • But isn't $q(A)$ an operator itself? Since $A$ is an endomorphism then $q(A)$ would result in another operator, so it wouldn't be a set, right? – Robert Lee May 31 '20 at 02:51
  • I never said q(A) is a set. The role of q(A) here in the context of the definition of kernel is the same as the element v, which q(A) certainly can be seen as one. – Divide1918 May 31 '20 at 02:56
  • I think I got confused with the notation. If I understood correctly, does the symbol "$\subset$" not mean subset in your answer? – Robert Lee May 31 '20 at 03:10
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    I fear there is a fundamental misunderstanding in this answer. Observe that $q(A)$ is an operator; however, $\ker(A - \lambda I)^r = {v \in V ,|, (A - \lambda I)^r(v) = 0},$ where $0$ in this case is the zero vector of $V.$ Hence, it does not make sense to say that $q(A)$ is contained in the kernel of the operator $(A - \lambda I)^r.$ – Dylan C. Beck Jun 03 '20 at 00:00
  • @Carlo You're right; the q(A) I wrote should actually be q(A)(v), where v is any vector in V – Divide1918 Jun 03 '20 at 10:29