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Question: A vector space is an abelian group with some extra structure. Given two vector spaces V1 × V2, show that the group V1 × V2 is a vector space.

Can someone explain to me the first sentence? Especially why it is commutative? As to the second sentence, is it just saying the product of vector spaces is vector space? Why it mentions "group" V1 × V2 particularly?

Thx in advance~

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    $\begingroup$ Part of the definition of a vector space is that it has an addition which is commutative (and makes it a group). Note that for this question to be correct, you need the two vectorspaces to be over the same field. $\endgroup$ Sep 13, 2013 at 12:44
  • $\begingroup$ Hi Tobias, thx for pointing it out, I do miss that part. $\endgroup$
    – Bob
    Sep 13, 2013 at 13:05

4 Answers 4

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As you will recall, a (real) vector space is a set $A$ together with operations ${+}: A\times A\to A$ and ${\cdot}:\mathbb R\times A\to A$ that satisfy certain conditions. It happens that those conditions imply that $(A,{+})$ is an abelian group. So if you take any vector space and forget how scalar multiplication works, what is left is an abelian group.

Conversely, you can also imagine that the vector space arose by taking an abelian group and then defining a scalar multiplication for it -- this is the sense in which one can say that a vector space "is an abelian group with additional structure"; the "additional structure" is the scalar multiplication. Note that it is not all abelian groups that can be made into vector spaces in this way, and some abelian groups have more than one possible scalar multiplication -- the same group can be made into a several different vector spaces by choosing different multiplication operations.

When the exercise says "the group $V_1\times V_2$", it is to underscore that the $\times$ there is a product of groups, so when you see $V_1\times V_2$ you only have a single operation, namely the group operation (which will become the addition in the vector space you're going to construct).

When it asks you to prove that $V_1\times V_2$ "is a vector space", it's using sloppy language. What it really means is to prove that the product group can be made into a vector space by finding an appropriate scalar multiplication $\mathbb R\times(V_1\times V_2)\to V_1\times V_2$ and showing that it satisfies the conditions for a vector space. It is up to you to puzzle out a scalar multiplication that will make this work; it doesn't already exist as a part of the group product $V_1\times V_2$.

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  • $\begingroup$ Can you give an example of an abelian group that, by adding the "additional structure", does not form a vector space. $\endgroup$ Mar 8, 2017 at 20:59
  • $\begingroup$ @theSongbird: The cyclic group with $4$ elements cannot become a vector space over any field. Neither can the group of nonzero reals under multiplication. $\endgroup$ Mar 8, 2017 at 21:14
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    $\begingroup$ I guess you'd prove that for $(R^*,/cdot )$ by assuming the opposite. $\endgroup$ Mar 9, 2017 at 13:26
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    $\begingroup$ @theSongbird: Yes. Namely: $-1$ is a group element of order $2$, so if there were a vector space structure on the group, then it would be a nonzero vector $v$ satisfying (in vector-space notation) $v+v=0$. This is only possible if $1+1=0$ in the scalar field, but if that is the case, then $v+v=0$ has to hold for all elements of the vector space, which is not the case for, say, $2$ -- a contradiction. $\endgroup$ Mar 9, 2017 at 13:31
  • $\begingroup$ But since your group is in $(R^*, /cdot)$, then from where are you getting the $+$ operation? $\endgroup$ Apr 1, 2017 at 16:56
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"A vector space is an abelian group with some extra structure." This follows from the axioms of a vector space. There are eight of them and the first four ones say exactly that $V$ is an abelian group:

  1. $\forall u,v,w\in V:\ u+(v+w)=(u+v)+w$,
  2. $\forall u\in V:\ u+0=u$,
  3. $\forall u\in V\ \exists -u\in V:\ u+(-u)=0$,
  4. $\forall u,v\in V:\ u+v=v+u$

1-3 say that $V$ is a group, 4 says it is abelian. Now what about the rest of axioms? They concern the multiplication by scalars and they are as follows:

  1. $\forall a\in\mathbb{K}\ \forall u,v\in V:\ a\cdot(u+v)=(a\cdot u)+(a\cdot v)$,
  2. $\forall a,b\in\mathbb{K}\ \forall u\in V:\ (a+b)\cdot u=(a\cdot u)+(b\cdot u)$,
  3. $\forall a,b\in\mathbb{K}\ \forall u\in V:\ a\cdot(b\cdot u)=(ab)\cdot u$,
  4. $\forall u\in V:\ 1_\mathbb{K}\cdot u=u$.

($1_\mathbb{K}$ is the multiplicative neutral element of the field $\mathbb{K}$). What do those axioms say? Let $End(V)$ be the ring of all endomorphisms of $V$ (ie. all group homomorphisms of $V$ into $V$). The multiplication in $End(V)$ is just a composition of functions and its neutral element is the identity function $id_V$: $id_V(u)=u$. Now what does it have to do with Axioms 5-8? They define a homomorphism from $\mathbb{K}$ into $End(V)$! How? Let $\varphi:\mathbb{K}\to End(V)$ be defined as follows $\varphi(a)(u)=a\cdot u$ where $a\in\mathbb{K}$ and $u\in V$. For each $a\in\mathbb{K}$ $\varphi(a)$ is a function from $V$ into $V$ -- it takes an element of $V$ and multiplies it by the scalar $a$. Axiom 5 say that for every $a\in\mathbb{K}$ $\varphi(a)$ is an endomorphism of $V$ ($\varphi(a)\in End(V)$), ie. $\varphi(a)$ is linear. Axioms 6-8 say that $\varphi$ is a homomorphisms between rings $\mathbb{K}$ (every field is a ring with multiplicative identity) and $End(V)$. This homomorphism $\varphi$ is the additional structure you are asking about.

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  • $\begingroup$ Is there a textbook you recommend which would have more information on this? Thanks! $\endgroup$ Jan 14, 2021 at 19:18
  • $\begingroup$ Usually this kind of approach is done in a pretty general way using so-called modules. You may want to check the chapter on modules and vector spaces in Hungerford's "Algebra" (or any other book on general algebra that also concerns modules). $\endgroup$ Feb 10, 2021 at 14:56
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In a vector space you can add vectors $v,w$ to get $v+w$, and one of the axioms is that $v+w = w+v$, so addition is commutative. You also have a zero vector $v+0=0+v=v$, so there is a group identity, and the other group properties also hold (inverse, associativity). However, in a vector space you can also multiply by a real number (and this multiplication has certain properties, like distributivity, etc.) which does not make sense if you only thought of the vector space as a group.

Defining vector spaces this way is quicker than defining then directly, because vector spaces have lots of axioms, many of which overlap with group axioms. The product of vector spaces is a group, because the product of groups is a group. So now you have a few more axioms to verify about $V_1 \times V_2$ to prove it is a vector space, but you don't need to waste time proving associativity or commutativity, which follow from group theory.

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Henning's answer is brilliant. But just to use some fancy language, a vector space is a more specific case of a module, in which the field of scalars has been replaced by a ring of scalars. Note that even a ring is always an additive abelian group, irrespective of how the multiplication looks. So the commutative nature of addition is sort of adapted in building a module and even a vector space.

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  • $\begingroup$ @anon Oh, right. $\endgroup$
    – Pedro
    Sep 14, 2013 at 21:09

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