# 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. Apr 2 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
Apr 2 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 Apr 2 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) Apr 2 at 21:29
• You ARE GIVEN addition and scalar multiplication in $F^n$. They are definitions 1.12 and 1.17 of the book. Apr 2 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? Apr 2 at 21:46