Amann/Escher, Analysis I, Remark 12.12: $K$-algebra homomorphism

Question

I'm reading reading Section I.12 Vector Spaces, Affine Spaces and Algebras from textbook Analysis I by Amann/Escher where there is Remark 12.12:

I would like to confirm if my understanding about $p(A) := \sum_{k} p_{k} A^{k}$ is correct.

1. $A^k = \underbrace{A \circ \cdots \circ A}_{k \text{ times}}$ where $\circ$ is function composition.

2. $p_{k} A^{k}$ is a function such that $(p_{k} A^{k}) (v) := p_k (A^k (v))$ for all $v \in V$.

3. $\sum_{k} p_{k} A^{k}$ is a function such that $\left ( \sum_{k} p_{k} A^{k} \right ) (v) := \sum_{k} \left [ ( p_{k} A^{k}) (v) \right ]$ for all $v \in V$.

Thank you for your help!

Answer 1

You are right. The only thing you have to know that on $\text{End}(V)$ you have an addition (pointwise by $(A + B)(v) = A(v) + B(v)$), a scalar multiplication (pointwise by $(k \cdot A)(v) = k \cdot A(v)$) and a multplication which is nothing else than function composition $(A \circ B)(v) = A(B(v))$. This makes $(\text{End}(V),+,\cdot,\circ)$ a $K$-algebra.

For any polynomial $p \in K[X]$ you may then insert any $A \in \text{End}(V)$ for the variable $X$. This generalizes to any $K$-Algebra $\mathfrak A$: $(p,\mathfrak a) \mapsto p(\mathfrak a)$ can be defined as for $\text{End}(V)$.

Answer 2

Because @Berci suggested a relevant question in his comment, I post my attempt here. It would be great if somebody helps me verify it.

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For $p,q \in K[X]$ and $v \in V$, we have

$$\begin{aligned} (pq) (A) (v) &= \left ( \sum_i \sum_{j \le i} p_j q_{i-j} X^i \right ) (A) (v) \\ &= \sum_i \sum_{j \le i} p_j q_{i-j} A^i (v) \\ & = \sum_{m} \sum_{n} p_m q_n A^{m+n} ( v) \quad (m:=j, n:=i-j)\end{aligned}$$

and

$$\begin{aligned} &(p(A) \circ q(A)) (v) \\ = & \left ( \sum_{m} p_m A^m \right ) \circ \left ( \sum_{n} q_n A^n \right ) (v) && = \left ( \sum_{m} p_m A^m \right ) \left ( \sum_{n} A^n (q_n v) \right ) \\ = & \sum_{m} p_m \left [ A^m \left ( \sum_{n} A^n (q_n v) \right ) \right ] && = \sum_{m} p_m \sum_{n} A^m (A^n (q_n v)) \\ = & \sum_{m} p_m \sum_{n} A^{m+n} (q_n v) &&= \sum_{m} \sum_{n} p_m q_n A^{m+n} (v)\end{aligned}$$

It follows that $(pq) (A) = p(A) \circ q(A)$ and thus the function $p \mapsto p(A)$ preserves multiplication.