# General proof for a Vector Space over F?

I've been self-studying Sheldon Axlers Linear Algebra Done Right and in the beginning of the book he covers the definition of a Vector Space. In one of his sentences he says "With the usual operations of addition and scalar multiplication, $$F^n$$ is is a vector space over $$F$$ as you should verify". But I'm confused as to how you would verify something like this?

What I know currently is that for a set to be a Vector Space (formally) it must satisfy a set of rules; commutativity, associativity, have an additive identity, additive inverse, multiplicative inverse and distributive properties. Additionally for a Vector Space V to be over a field F is must satisfy scalar multiplication and vector addition. Would we prove the second then in this case? In the example with $$F^{\infty}$$ Axler provides he defines these then still asks us to verify that $$F^{\infty}$$ is a vector space over $$F$$ which confuses me more. I would like to understand this before moving on, any help would be appreciated.

• You define the operations; but the definition of the operations, in and of itself, does not prove that they satisfy the required properties. You must prove that they do. Commented Apr 2, 2021 at 21:26
• @ArturoMagidin What properties have to be satisfied? The ones I listed? And what about in the case where no operations are defined? Do we assume they are?
– kman
Commented Apr 2, 2021 at 21:28
• "$V$ is a vector space" is just short "$V$ is a vector space over the field $F$". Just some times you omit the field since it is understood from the context. They are not two different concepts Commented Apr 2, 2021 at 21:28
• For example, say I want to define a vector space over $\mathbb{R}$ using the positive reals as vectors. I need to tell you how to do "vector addition" of these "vectors", and how to do scalar multiplication. I tell you: "to 'vector-add' to positive reals, you multiply them: $a\oplus b= ab$, where $\oplus$ is the vector addition I am defining, and the right hand side is the usual product. To 'scalar-multiply' a real number by a positive real, you take the exponent: $r\odot a = a^r$, where $\odot$ is the scalar multiplication I am defining." (cont) Commented Apr 2, 2021 at 21:29
• You ARE GIVEN addition and scalar multiplication in $F^n$. They are definitions 1.12 and 1.17 of the book. Commented Apr 2, 2021 at 21:38

The first part (abelian group - or in your terms: commutativity, associativity, have an additive identity, additive inverse) holds for $$F^n$$ (as well as for $$F^\infty$$) by defining the addition in an obvious manner, namely componentwise.
The second part, (the field action - or in your terms: distributive properties and multiplicative inverse [though I'd prefer that $$1\in F$$ acts as identity]) also follows by defining the scalar multiplication in the obvious manner, i.e., component-wise.
You may also notice that $$F^n$$ and $$F^\infty$$ can be viewed as the set $$F^X$$ of all maps from a certain set $$X$$ to $$F$$ (e.g., take $$X=\{1,\ldots,n\}$$ to obtain $$F^n$$). In this view, component-wise addition/multiplication just means point-wise addition/multiplication, i.e., $$f+g$$ is the map $$X\to F$$ given by $$(f+g)(x)=f(x)+g(x)$$ and $$a\cdot f$$ is the map $$X\to F$$ given by $$(a\cdot f)(x)=a\cdot f(x)$$. The desired laws then immediately follow from their validity for addition and multiplication in $$F$$ itself, i.e., because $$F$$ is a field.
• The OP is struggling with the fact that $F^n$ is a vector space and you talk about an action of a field over an abelian group and give an more abstract definition of $F^n$? Really? Commented Apr 2, 2021 at 21:46