0
$\begingroup$

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.

$\endgroup$
13
  • $\begingroup$ 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. $\endgroup$ Commented Apr 2, 2021 at 21:26
  • $\begingroup$ @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? $\endgroup$
    – kman
    Commented Apr 2, 2021 at 21:28
  • $\begingroup$ "$V$ 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 $\endgroup$
    – jjagmath
    Commented Apr 2, 2021 at 21:28
  • $\begingroup$ 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) $\endgroup$ Commented Apr 2, 2021 at 21:29
  • 2
    $\begingroup$ You ARE GIVEN addition and scalar multiplication in $F^n$. They are definitions 1.12 and 1.17 of the book. $\endgroup$
    – jjagmath
    Commented Apr 2, 2021 at 21:38

1 Answer 1

0
$\begingroup$

If you do linear algebra really right, you define a vector space with a single axiom as

an abelian group on which a field acts.

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.

$\endgroup$
1
  • 4
    $\begingroup$ 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? $\endgroup$
    – jjagmath
    Commented Apr 2, 2021 at 21:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .