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As per the definition of Vector space, for a set of vectors and scalars, the only operations allowed are the 2 closure properties(for addition and scalar multiplication) and 10 axioms(for associativity commutativity, inverse and identity) for the vectors, scalars involved. If so, where does dot product and cross product(which is vector time vector multiplication) fit in? These 2 vector times vector multiplication operations are not defined and hence not allowed as per the definition of Vector space. Is this understanding correct or am I missing something here?

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  • $\begingroup$ Ultimately they're just functions from $V \times V$ (where $V$ is your vector space) to your scalar field or into the vector space (dot, cross product respectively). While the definition of vector space gives us a means to characterize sets meeting certain properties under a couple of special functions (operations), we have to make further definitions to get the more general notions of dot/cross products, the inner and outer products. But these are just, again, special functions meeting certain criteria. $\endgroup$ Jan 10, 2021 at 8:12
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    $\begingroup$ The basic abstract definition of a vector space does not include any notion of distance. Vector spaces, indeed, can be formed with any field providing the scalars. But one can study quadratic forms, bilinear forms, tensor products and the like and generalise the notions of product. It is not that products are "not allowed" - but where they are defined they provide additional structure. The less structure a mathematical system has, the more generally it can be applied, Vector spaces without defined products turn out to be very useful. $\endgroup$ Jan 10, 2021 at 8:28

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It depends on your structure. For example, I prefer to define vector space with a single axiom

A vector space is an abelien group $V$ on which a field $k$ acts

Let's look at this for the vector space $V=\Bbb R^3$. Initially, we can see that $\Bbb R^3$ is an abelien group under addition; this is so because of course $\Bbb R$ is an abelien group under addition and for every abelien group $A$, we can naturally define an abelien group structure on $A\times A$ or $A\times A\times A$ or ... component-wise. Viewed as an abelien group, we only have the addition operation at our disposal. Of course $\Bbb R$ itself is much more than an abelien group. We might view $\Bbb R^3$ as a non-unitary ring by also defining multiplication component-wise from the underlying ring $\Bbb R$. But that is of no interest when we want to have a vector space. The ring $\Bbb R$ happens to be a field, but that property does not transfer to a product. And $\Bbb R$ is also a topological space (or even a metric space), which does allow us to define a natural structure of topological space on $\Bbb R^3$. This does not give us a new (binary) operation on $\Bbb R^3$, but allows us to speak og convergence and limits. In fact, the topology is compatible with the group law, i.e., $\Bbb R^3$ is a topological abelien group. But in the context of vector space, we care only for the abelien group structure.

As with every abelien group, the set $\operatorname{End}(V)$ of endomorphisms $A\colon V\to V$ is endowed with a natural ring structure by defining $(A+B)(v)=A(v)+B(v)$ and $(A\cdot B)(v)=A(B(v))$. In particular, the identity endomorphism is the $1$-element of this ring. To make $\Bbb R^3$ a vector space, we need a base field (and decide to use $k=\Bbb R$) and an action of $k$, i.e., a ring homomorphism $\phi\colon k\to \operatorname{End}(V)$ and for convenience write $x\cdot v$ for $\phi(x)(v)$, giving us a new operation $\cdot\colon k\times V\to V$ that automatically has some nice properties: E.g., $1\cdot v=v$ follows because in a homomorphism between unitary rings, we postulate that the 1-element is mapped to the 1-element. And things like $(x+y)\cdot v=x\cdot v+y\cdot v$ or $x\cot(v+w)=x\cdot v+x\cdot w$ or $(x\cdot y)\cdot v = x\cdot(y\cdot v))$ also follow immediately from the definition of ring homomorphism. Everything acts together so nicely that using the same operation symbols $+$ and $\cdot$ for various different operations does not confuse us (and in fact is so suggestive that it simplifies our life a lot).

But how do we actually get such a ring homomorphism? There is one natural way, using the fact that $V$ is a direct product of copies of $k$: We map $x\in k$ to the endomorphism of $k^3$ that multiplies component-wise by $x$ (using the fact that each factor $k$ is our field). This is how we could define a (natural) structure of a real vector space on $\Bbb R^3$, arriving at the usual vector space operations. What we do not have in general is additional structure beyond that of a vector space.

I already mentioned that $\Bbb R^3$ is a topological group. All operations including multiplication with a scalar are continuous under this topology and that allows us to view $\Bbb R^3$ as something more than just a vector space: It is a topological vector space. Not every vector space is also a topological vector space.

I also noted that we can consider component-wise multiplication to view $\Bbb R^3$ as a (non-unitary) ring; this operation is also compatible with the other operations and allows us to view $\Bbb R^3$ as a (non-unitary) algebra over $\Bbb R$. But we cannot do the same for every vector space.

Next, in our specific example we have a nice bilinear map $V\times V\to \Bbb R$ given by multiplying component-wise and adding the results - the dot product. This turns $\Bbb R^3$ into an inner product space. Again, not every vector space is an inner product space, and even if we have an inner product, it is not uniquely determined. But in the case of $\Bbb R^3$ (or more generally, $\Bbb R^n$), we have a natural way to define an inner product. But to repeat: A vector space is in general not endowed (or even endowable) with such an inner product.

Finally, the cross product. To give the broader picture about this, I'd have do dive quite deeply into abstract structures (even compared to what I have done so far). Let's leave it at this: Just as not every vector space is a topological space, or an algebra, or an inner product space, it is also not the case that every vector space allows a cross product. To keep things simple, you may rest assured for the moment that this is only available in a natural way for the specific vector space $\Bbb R^3$. In particular, there is not even a cross product in $\Bbb R^2$ or $\Bbb R^4$.

Summary: If be begin with "Let $V$ be a real vector space ..." then all we have are the vector space operations and are not "allowed" to use dot product or cross product. If we begin "Let $V$ be an inner product space ..." we have the dot product at our disposal. This is also implicitly the case if be begin "Let $V$ be the vector space $\Bbb R^n$ for some $n\in\Bbb N$ ...". But we can use the cross product only in a context of "Let $V=\Bbb R^3$ ...".

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  • $\begingroup$ Nice answer. I'm guessing it is too abstract for the OP, so I am providing my own (more mundane) answer. $\endgroup$
    – GEdgar
    Jan 10, 2021 at 10:15
  • $\begingroup$ Have read through your reply. The summary is very clear and clarified my doubt. Thanks. $\endgroup$ Jan 18, 2021 at 1:48
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You are correct, an abstract vector space (assume all spaces I mention have $\mathbb R$ as scaars) does not have such operations (dot product, cross product) defined.

An "inner product" on a vector space $V$ is an operation with the properties of a dot product. Any abstract vector space admits many different inner products. A vector space, together with a specified inner product, is called an inner product space . This name exists because mathematicians have found this type of structure useful.

An unexpected result says that an operation like "cross product" only exists for $3$-dimensional spaces. There is an "analog" of the cross product in $\mathbb R^n$, but in that case you multiply $n-1$ vectors to get a vector answer. So only the case $\mathbb R^3$ is where you multiply two vectors to get a vector answer. See HERE for various generalizations of the cross product to other vector spaces.

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  • $\begingroup$ The inner product space makes it clear. Thanks for responding to the question. $\endgroup$ Jan 18, 2021 at 1:49

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