In the book "Linear algebra done Right" p.14 the example problem asks to prove that if S is non-empty then $F^S$ is a vector space. Addition is defined as $(f+g)(x) = f(x) + g(x)$ and scalar multiplication is defined as $(af)(x) = a*f(x)$ for all $a \in F$ and $f,g \in F^S$
What if S is empty? It seems to me that $F^S$ could still be a vector space. Do we need to change the definitions of addition and scalar multiplication to make this work?
If we change the definitions of addition and multiplication I think we can make it work as follows:
Suppose S is empty. Then there is only one function $f:S \rightarrow F$, namely the empty function $\varnothing$. Define vector addition as $u + v = \varnothing$ for all $u,v \in F^{\varnothing}$. Then define scalar multiplication as $av = \varnothing$. Are these definitions okay? I am sorry to use $\varnothing$ for both the empty set and the empty function (I couldn't find any other symbol commonly used for the empty function).
With these definitions it seems to me that $F^{\varnothing}$ is commutative and associative. The additive identity could be the empty function, making the additive inverse also the empty function. The multiplicative identity works since 1 multiplied by the empty function is defined as the empty function. I believe the distributive properties hold as well. Does this work?
Could $F^S$ be a vector space with the usual definitions of addition $(f+g)(x) = f(x) + g(x)$ and scalar multiplication $(af)(x) = a*f(x)$ for all $a \in F$ and $f,g \in F^S$. Maybe I do not understand the empty function enough to be able to tell...
Please advise
Your understanding is quite right. $F^\emptyset$ comprises a single element that we can call $0$ and take as the unique element of a zero-dimensional vector space over $F$. I suspect the authors of "Linear algebra done right" just didn't want to distract their readers into worrying about this special case: you quite rightly noted the special case and drew the correct conclusion.