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

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 $$v \in V$$.

• Yes, correct. Also, $A^0$ is the identity, so that $A^0v=v$. Do you see why is this mapping $p\mapsto p(A)$ preserves multiplication? Jul 5 '19 at 8:48
• Hi @Berci, I have posted my answer for your question below. Could you please verify it? Thank you so much! Jul 6 '19 at 13:41

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 $$\text{End}(V)$$.

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

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.

• Yes, that's correct. Jul 6 '19 at 13:44
• Thank you so much for your support @Berci ;) Jul 6 '19 at 13:45
• Actually, by linearity it suffices to verify that $x^n$ is mapped to $A^n$.. Jul 6 '19 at 13:56
• Hi @Berci, I think that "$x^n$ is mapped to $A^n$" is quite obvious from $p \mapsto p(A)$. Jul 6 '19 at 14:02