Tensor product of vector space - question about definition We were given the following definition of the tensor product of vector spaces, and I don't quite understand what this is actually supposed to tell me: 
Definition:
Assume $K$ is a field and $V_1, ..., V_n$ are vector spaces over $K$. Let $F$ be the vector space over $K$ that is constructed by every symbol $(v_1, ..., v_n)$ with $v_i \in V_i$. We notate the elements of the base as $e_{(v_1, ...., v_n)}$. Assume $U \subset F$ is the vector subspace over $K$ that is constructed by every element that posesses the form 


*

*$r\cdot e_{(v_1, ..., v_{i-1}, v_i, v_{i+1}, ..., v_n)} - e_{(v_1, ..., v_{i-1}, r\cdot v_i, v_{i+1}, ..., v_n)}$

*$e_{(v_1, ..., v_{i-1}, u+w, v_{i+1}, ..., v_n)} - e_{(v_1, ..., v_{i-1}, u, v_{i+1}, ..., v_n)} -  e_{(v_1, ..., v_{i-1}, w, v_{i+1}, ..., v_n)}$
Then we call $F/U$ (quotient space) the tensor product of the $V_i, i \in \{1, ..., n\}$.
Questions:


*

*Why is he talking about "symbols"? Aren't these just regular vectors? 

*What does $e_{(v_1, ...., v_n)}$ mean? It came out of nowhere here. I guess it has something to do with the canonical vectors, but how do I have to deal with this kind of notation? 

*What doesn't make sense to me either is the definition of $U$. How is this actually defining a vector space? 
 A: To understand the definition you should first understand the universality of tensor product.
Motivation


*

*Let $U, V$ be vector spaces over $K$. We can build a multiplication of elements of $U, V$ in values in some vector space $W$, more formally a map
$$U\times V\to W,\quad (u,v)\mapsto u\star v$$
that satisfies the natural properties of multiplication -- distributivity and associativity, i.e.
$$(u_1+u_2)\star v=u_1 \star v+u_2\star v,\\
u\star(v_1+v_2)=u\star v_1+u\star u_2,\\
k(u\star v)=(ku)\star v=u\star (kv),\quad k\in K.$$

*Then, if we have a homomorphism $f:W\to W'$ we automatically have defined a multiplication with values $W'$, namely
$$U\times V\to W',\quad (u,v)\mapsto f(u\star v)$$
that also satisfies the distributivity and associativity laws.

*Ask a question: is there a universal space $W_\mathrm{univ}$ such that any other multiplication $U\times V\to W$ can be defined as in the item 2, namely $(u,v)\mapsto f(u\star_\mathrm{univ} v)$ where $f:W_\mathrm{univ}\to W$ is a homomorphism? The answer is yes. Such $W_\mathrm{univ}$ is called the tensor product of $U,V$ denoted as $U\otimes_K V$, and $\star_\mathrm{univ}$ denoted as $\otimes$. The homomorphism $f$ is always exists and unique!
In other words, tensor product provides universal multiplication of elements of given vector spaces. The only restrictions that it holds is  distributivity and associativity laws.
Construction of the tensor product
Let $u_1,\ldots,u_n$ be a basis of $U$, $v_1\ldots,v_n$ be a basis of $V$. Then consider the vector space $M$ of formal linear combinations of symbols $(u_i, v_j)$ (in the notation of your book it's $e_{u_i,v_j}$; but no matter how you notate it, they are just symbols for basis vectors of $M$). For example the $4(u_2,v_1)-2(u_5,v_2)+42(u_3,v_7)$ is an element of $M$ (of course, if each mentioned scalar is in $K$ and each mentioned $u_i,v_j$ is in resp. $U,V$). Check that it's a vector space.
Then we want to satisfy distributivity and associativity law. To do that just define some combinations to zero:
$$k(u_i,v_j)-(ku_i,v_j)=0,\\
(u_i+u_j,v)-(u_i,v)-(u_j,v)=0,\\
\ldots\text{and so on}$$
More formally we consider the space $M_0$ containing such combinations, and take the quotient space $M/M_0$, that is all formal linear combinations of $(u_i,v_j)$ up to combinations like $k(u_i,v_j)-(ku_i,v_j)$ which defined zero. $M/M_0=U\otimes_K V$.

I used only two spaces $U$ and $V$, but of course all holds for any finite number of vector spaces.
See more formal details in your book, Lang "Algebra", or Kostrikin, Manin "Linear Algebra and Geometry", for example.
A: (1) It's true that the set of $n$-tuples $(v_1,\cdots,v_n)$ carries a vector space structure with compontwise operations. However, we're completely ignoring that operation. We're only thinking of these tuples as elements of the set $V_1\times\cdots \times V_n$ as a Cartesian product of sets. In order to emphasize the fact we are treating them as elements of a set we're going to call them "symbols" instead of "vectors," since (again) we are ignoring the usual vector space structure on $V_1\times\cdots\times V_n$.
(2) If $X$ is a set, then the free vector space (over $K$) generated by $X$ could have its elements denoted by linear combinations of elements of $X$, like $3x_1-x_2$ (with $x_1,x_2\in X$) for instance, but this gets weird if $X$ already has operations that we're ignoring that might confuse us. As an example, consider the free vector space generated by the set $X=\{1,2,3\}$. If we did the usual linear combination thing, an element might look like $3(1)-(2)$, where we're treating $(1)$ and $(2)$ as elements of $X$ instead of as numbers. This is very confusing. The fix is to relabel all the elements $x\in X$ as "basis vectors" $e_x$ instead, so we could write $3e_1-e_2$ and $\{e_1,e_2,e_3\}$ would be a basis.
The set $V_1\times\cdots \times V_n$ already has additive and multiplicative structure - the usual operations would tell us for instance that
$$r(v_1,\cdots,v_n)=(rv_1,\cdots,v_n) \\ (v_1,\cdots,v_n)+(w_1,\cdots,w_n)=(v_1+w_1,\cdots,v_n+w_n) $$
These relations are irrelevant in the free vector space on the set $V_1\times\cdots\times V_n$, since (because of the "free" part) we require every single tuple to be linearly independent. The above relations would be false, and therefore operations would be very confusing if we just kept using tuples. Instead, we'll put the tuples down below a letter "e" as subscripts.
(3) It describes a large subset of the free vector space, and $U$ is the subspace spanned by these elements. Taking the linear span of a subset by definition yields a vector subspace.

Formal definitions can be very tedious, and require clever algebraic tricks to accomplish intuitive tasks. Just look at the formal construction of integers, rationals, and reals all from the naturals, or the formal construction of the naturals using the empty set recursively. But there's usually some kind of intuitive concept that is guiding the design of the formal construction.
It's perhaps easier to work with two vector spaces at a time, say $V$ and $W$. Suppose we want to "pretend multiply" elements $v\in V$ with $w\in W$. Well, we'll need a symbol for our pretend multiplication, so let's call it $\otimes$. Thus, we end up crafting a collection of things that look like $v\otimes w$ for various vectors $v\in V,w\in W$. (This corresponds to thinking of $V\times W$ as a set of labels.)
Next, we want these pretend products to reside within a new vector space, where we have addition and scalar multiplication, so by the closure property we're going to need linear combinations of things that look like $v\otimes w$, such as
$$ a_1(v_1\otimes w_1)+ a_2(v_2\otimes w_2)+\cdots $$
(This corresponds, in our formal construction, to taking the free vector space generated by the set $V\times W$.) For the moment, don't worry about what any of these things mean. We're only playing pretend, after all, so there's no reason to assume this means anything until we create an interpretation later.
Anyway, another thing we may desire to have. If $\otimes$ is indeed to represent a kind of multiplication operation, then we should require it satisfy the usual distributive property,
$$ (v_1+v_2)\otimes w=v_1\otimes w+v_2\otimes w \\ v\otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2 $$
We could also want to scalar multiply the symbols $v\otimes w$ in a way that interacts with the usual scalar multiplication on $v$ and $w$, as in
$$r(v\otimes w)=(rv)\otimes w=v\otimes (rw). $$
These relations are not a priori true, since formally we're dealing with a free vector space in which these symbols $v\otimes w$ are all linearly independent. But is there anyway we could pretend they are true, and every logical consequence of them is true, and be talking about something real? Yes! In the above equalities, subtract over to one side to get (stuff) = 0, then take the linear span of all of the things that appear on the left-hand side of these equalities, then quotient by that vector subspace. In the resulting quotient space, all of these relations will be true!
As an analogy, if we wanted a ring with an element $x$ satisfying $x^2=x+1$, we could take the free construction (polynommial ring) $\mathbb{Z}[x]$, then quotient by the ideal generated by $x^2-x-1$ to get $R=\mathbb{Z}[x]/(x^2-x-1)$. By taking relations, turning them into (stuff)=0 (or the analogous versions of that equality), generaing a subobject with them, and computing the quotient, we are imposing the relations and all of their logical consequences.

Now, $\otimes$ is not really a multiplication operation on $V\otimes W$. For one thing, it takes vectors from $V$ on the left, vectors from $W$ on the right, and returns vectors from $V\otimes W$ as outputs. In an actual multiplication map, we would want some kind of closure under the operation.
Why would we want to construct such a thing in the first place? The main use is in converting multilinear algebra to regular old linear algebra. For simplicity, let's keep working with two vector spaces $V$ and $W$, so let $f:V\times W\to Z$ be any bilinear map, where $Z$ is another vector space. (Note it is not a linear map on $V\times W$ with the componentwise operations.)
To be bilinear, it must satisfy the relations
$$ f(v_1+v_2,w)=f(v_1,w)+f(v_2,w) \\ f(v,w_1+w_2)=f(v,w_1)+f(v,w_2) \\ rf(v,w)=f(rv,w)=f(v,rw). $$
These should look familiar! If we just drop the $f$s, these are exactly the relations we were trying to impose on the free vector space on the elements of $V\times W$ (although we used letter $e$s or $\otimes$ symbols above). Indeed, reinterpreting the above relations in that light, they say that $f$ is linear as a function of elements of the tensor product.
That is, we have the following "universal property": for any bilinear map $f:V\times W\to Z$, there is a unique linear map $\tilde{f}:V\otimes W\to Z$ for which $f$ is the composition $V\times W\to V\otimes W\to Z$.
The tensor product also comes in handy in other ways too. For instance, $\hom(A,B)$ (the vector space of linear maps $A\to B$) is naturally equivalent to the tensor product $A^*\otimes B$.
