# function spaces in linear algebra

I am in upper division linear algebra, and i need help in proving a function space as a vector space. I just need help proving 2 particular vector space axioms.

Axiom 1: For each pair of elements a,b in F(field), and each element x in V (Vector space), ab(x)= a(bx).

Axiom 2:For each element a in F and each pair of elements x,y in V a(x+y)=ax+ay

To prove axiom 1, you would need 2 scalars, so can I just say, let a and b be a member of F... So wouldn't the proof be exactly the same as the axiom? ab(s)= a(b(s))?

To prove axiom 2: I really don't know where to start....

What are your tips and tricks to prove these axioms when you have function spaces? I am having a lot of trouble proving how the axioms work. Any help would be greatly appreciated...

• Proving an axiom...? – IAmNoOne Sep 27 '16 at 23:21
• en.wikipedia.org/wiki/Axiom#Non-logical_axioms – Stanley Sep 27 '16 at 23:27
• @Stanley, never knew that. – IAmNoOne Sep 27 '16 at 23:58
• Yeah it's probably a misnomer. But it's very common to talk about group axioms, vector space axioms, etc. – Stanley Sep 28 '16 at 0:02

It's generally a bad idea to try and work with a vector space when you haven't got a good idea of how it's defined. What do we mean by scalar multiplication? Addition of two vectors? The way it's usually set up is as follows:

If $V$ is a set of functions from $X$ to $Y$, where $Y$ is a vector space over a field $F$, then we define the addition of two functions:

$(f+g)(x) = f(x) + g(x)$ for all $f, g \in V$ and all $x \in X$.

and multiplication by scalars:

$(\alpha f )(x) = \alpha(f(x))$ for all $f \in V, \alpha \in F$ and $x \in X$.

Once you have understood the setup, it's pretty easy to show the axioms are fulfilled.

• I find it interesting that we don't need $X$ to be a vector space, or have any algebraic structure at all :) Maybe that's just me. – Stanley Sep 27 '16 at 23:08
• Hi Stan, Yes, that is what it states in my book and notes, but I am still having trouble coming up with a strategy to prove the axioms . – user6873843 Sep 27 '16 at 23:08
• @Bye_World has a more complete answer. – Stanley Sep 27 '16 at 23:11
• Yes, indeed. It might help to put names to this construction: pointwise addition and scalar product. It is possible to define these differently, in such a way that they do not satisfy the vector space axioms, which goes to show that the question is incomplete. The question-poser was either given this structure, or you are expected to deduce it as the most "natural" way in which to define these. – qman Sep 27 '16 at 23:20
• For axiom 2, $[a(f+g)](x) = a[(f+g)(x)] = a[f(x)+g(x)] = af(x) + ag(x) = (af)(x) + (ag)(x)$, so we conclude that $a(f+g) = af + ag$. I've used notations $f$ and $g$ to denote the functions, $x \in X$ and $a \in F$. – Stanley Sep 27 '16 at 23:22