2
$\begingroup$

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:

enter image description here

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!

$\endgroup$
2
  • $\begingroup$ 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? $\endgroup$
    – Berci
    Jul 5 '19 at 8:48
  • $\begingroup$ Hi @Berci, I have posted my answer for your question below. Could you please verify it? Thank you so much! $\endgroup$
    – Akira
    Jul 6 '19 at 13:41
2
$\begingroup$

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 $\text{End}(V)$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
4
  • 1
    $\begingroup$ Yes, that's correct. $\endgroup$
    – Berci
    Jul 6 '19 at 13:44
  • $\begingroup$ Thank you so much for your support @Berci ;) $\endgroup$
    – Akira
    Jul 6 '19 at 13:45
  • $\begingroup$ Actually, by linearity it suffices to verify that $x^n$ is mapped to $A^n$.. $\endgroup$
    – Berci
    Jul 6 '19 at 13:56
  • $\begingroup$ Hi @Berci, I think that "$x^n$ is mapped to $A^n$" is quite obvious from $p \mapsto p(A)$. $\endgroup$
    – Akira
    Jul 6 '19 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.