I know that a vector space is over a field and has the operations of addition (among elements, which are called vectors) and scalar multiplication (in which a field element acts on a vector to yield a vector). Now, I want to reconcile this understanding with my previous intuition by asking some questions.

  1. Why can't a vector space have cross products/dot products defined inside it? What problems would be run into? In general, what kinds of "products" can we define in a vector space?

  2. Over what fields is $\mathbf{R}^n$ considered a vector space?

  3. Is it correct that vector space isomorphism only exists in the same field? Or can we claim that different vector spaces in different fields are isomorphic? What is the intuition used to extend vector spaces to infinite dimensions in functional analysis?

  4. Why do we only have to check for closure under addition and scalar multiplication to prove that a set is a vector space when a vector space has six (or seven, depending on how you count) axioms?

I'm studying linear algebra for the second time in a proof-based context, so answers relevant to this level would be appreciated!

  • $\begingroup$ 1. a vector space can have products defined inside it but doesn't have to; the product could be a scalar (inner product space) or a vector (algebra) $\endgroup$ Feb 4, 2021 at 0:19
  • $\begingroup$ In $4$ you're looking at a subset of given vector space, so you already know that addition is commutative and associative, etc. What you need to check is closure. (Once you have closure of addition and scalar multiplication, the existence of additive inverses follows.) $\endgroup$
    – saulspatz
    Feb 4, 2021 at 0:24
  • $\begingroup$ @saulspatz I agree with your answer if the problem gives us a vector space and asks us to identify whether a certain set is a subspace. A clarification of my question is as follows: To prove that a set is a real vector space, is all we need to check closure and scalar multiplication, or do we need to check all axioms? In other words, does closure under addition and scalar multiplication from elements of $\mathbb{R}$ imply all of the axioms in the definition of a vector space or is it necessary to check all of the axioms one-by-one to see if they hold? $\endgroup$
    – user404974
    Feb 4, 2021 at 23:45
  • $\begingroup$ @ExponentBOT If it's just an arbitrary set, you have to show that each of the axioms holds. $\endgroup$
    – saulspatz
    Feb 5, 2021 at 0:37

2 Answers 2


Question 1.

A general definition of a "product" is the following: suppose you have three vector spaces $X,Y,Z$ all over the same field. Then, a bilinear mapping $\beta:X\times Y\to Z$ is essentially what we mean by a "product". So you can certainly consider the notion of products/multiplication. But, what is important to note is that this is extra information that you have to provide; it is not part of the vector space axioms, hence there is no standard/canonical choice in general.

You certainly can look at vector spaces equipped with dot products (more commonly called inner products). More precisely, suppose $V$ is a vector space over $\Bbb{R}$. Then, an inner product on $V$ is a bilinear, symmetric positive-definite mapping $\langle\cdot,\cdot\rangle : V\times V \to \Bbb{R}$. By prescribing an inner product, we are in essence prescribing a geometry for the vector space $V$. If you want to generalize some more, we can weaken the conditions imposed. For instance, a bilinear, symmetric, non-degenerate mapping $g:V\times V\to \Bbb{R}$ is called a pseudo-inner product, and we refer to the pair $(V,g)$ as a pseudo-inner product space. (I intentionally left out the complex case since the definitions are slightly different). The notion of a cross product is more subtle so let me not go into the details; but really if you formulate an appropriate definition, you can define various types of "products".

Question 2.

If $\mathbb{F}$ is any subfield of $\Bbb{R}$, then we can consider $\Bbb{R}^n$ as a vector space over $\Bbb{F}$ (by restricting the "usual" operations to the field $\Bbb{F}$). The most obvious examples are $\Bbb{F}=\Bbb{R},\Bbb{Q}$.

Question 3.

Suppose $V,W$ are vector spaces over fields $F_1,F_2$ respectively. To define the notion of a linear map $T:V\to W$, we would like the following equation to hold for all $x,y\in V,\lambda\in F_1$: $T(\lambda x+y)=\lambda T(x)+T(y)$. Well, to talk about $\lambda T(x)$, we need the target space to be considered as a vector space over $F_1$, so we'd need $F_1\subset F_2$. If you want to talk about isomorphisms, then you would also need the inverse map to be linear, so we want $F_2\subset F_1$. Thus, we always consider them over the same field $F$.

Question 4.

If $V$ is a vector space over a field $F$, and $W\subset V$ is any non-empty subset, then of course, by definition, to say $W$ is a vector space over $F$ we have to verify a whole bunch of axioms. But, it's a nice fact that we actually don't have to do so much work: as long as we check $0\in W$ (or just that $W\neq \emptyset$), and that $W$ is closed under addition and scalar multiplication, then it follows that $W$ (equipped with the restricted addition and scalar multiplication as operations) is a vector space over $F$ in its own right (this is usually called the "subspace" criterion or something). The nice thing is all the axioms like associativity/commutativity or addition/ distributive laws etc all hold in $W$ because they hold in $V$ and since $W\subset V$.

  • $\begingroup$ @Marktmeister sure, I agree with that and thanks for emphasizing that (though I don't think what I wrote suggests anything to the contrary; I did say "we need the inverse map to be linear..." so that the very notion of linearity $W\to V$ requires $F_2\subset F_1$) $\endgroup$
    – peek-a-boo
    Feb 4, 2021 at 10:57
  • $\begingroup$ @Marktmeister OP asked "Is it correct that vector space isomorphism only exists in the same field", and my answer is yes we must consider it over the same field, and I tried to explain why: if $V,W$ are vector spaces over certain fields and we have a bijective $T:V\to W$, then to say that $T$ is an isomorphism means $T$ is linear, (hence the requirement $F_1\subset F_2$) and we also need by definition $T^{-1}:W\to V$ to be linear and hence we must require $F_2\subset F_1$. I feel like now we're going off on a tangent regarding a very minute phrasing issue (which I don't see tbh). $\endgroup$
    – peek-a-boo
    Feb 4, 2021 at 11:15
  • $\begingroup$ Ok, I misread. I'm deleting my comments. $\endgroup$ Feb 4, 2021 at 11:49
  • $\begingroup$ @peek-a-boo So, my interpretation of your answer to #3 is that vector space isomorphism only exists in the same field as a consequence of the definition of isomorphism as a bilinear mapping. Is this accurate? $\endgroup$
    – user404974
    Feb 5, 2021 at 3:17
  • $\begingroup$ @Exponent BOT careful: the word "bilinear" has its own technical meaning and its not what (I think) you intended. but yea it's because we require the function and its inverse to both be linear. $\endgroup$
    – peek-a-boo
    Feb 6, 2021 at 12:31
  1. You can define an inner product on every finite dimensional real vector space, in fact if $V$ is a n-dimensional vector space it is isomorphic to $R^n$ and you can use this isomorphism to define an inner product. This definition is not canonic(it depends on the basis you choose). Cross product is defined only on $R^3$, you could define a generalization also in $R^n$ but it is not so simple(see Hodge star operator).

  2. $\mathbb{R}^n$ is normally considered as a $\mathbb{R}$ vector space. Of course it can also be equipped with a structure of vector space over other fields(for example over $\mathbb{Q}$), but it is not the standard structure.

  3. Because morphisms of vector spaces over different fields cannot be defined. Consider $V$ and $W$ vecotr spaces over $F$ and $E$. Let $f:V\rightarrow W$ be a linear map. Let $\lambda\in F$. The axiom $f(\lambda x)=\lambda f(x)$ doesn't make sense, in fact $\lambda \notin E$ and so it is not defined $\lambda f(x)$.

  4. Because if $V$ is a vector space and $W$ is a subset of $V$, commutativity, associativity etc. are satisfied because elements of W are also elements of V

  • $\begingroup$ What does $K^n$ represent? $\endgroup$
    – user404974
    Feb 5, 2021 at 3:23
  • $\begingroup$ Sorry, I meant $R^n$. $\endgroup$
    – Luca.b
    Feb 5, 2021 at 10:30

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