Suppose 𝑆 is a nonempty set. Let $𝑉^𝑆$ denote the set of functions from 𝑆 to 𝑉. Define a natural addition and scalar multiplication on $𝑉^𝑆$, and show that $𝑉^𝑆$ is a vector space with these definitions.

so to start by defining natural addition

$(f+g)(x)= f(x)+g(x)$


$(\lambda g)(x)= \lambda g(x)$

the conditions a vector space must satisfy are

  1. commutativity

$f(x)+g(x)= g(x)+f(x)$ by the natural definition of addition

  1. associativity

let's define 3 functions $q(x), p(x), r(x)$

by the definition of addition $(q(x)+p(x))+ r(x)=q(x)+(p(x)+r(x))$

3)the additive identity

let $f(x)=0$

then we have $0+g(x)=g(x)$

  1. additive inverse

let $f(x)=-g(x)$

then we have $f(x)+g(x)=0$

  1. Multiplicative identity

let $g(x)=1$

then we have $g(x)f(x)=f(x)$

the distributive property follows similarly

this is problem 7 in exercise 1B of axler 4e

thus $V^S$ is a vector space is that right?

  • 1
    $\begingroup$ Yes, nothing wrong here. Here you can find solutions to the exercises in the book. $\endgroup$
    – Bowei Tang
    Commented Jun 11 at 12:06


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