The definition I have of a tensor product of vector finite dimensional vector spaces $V,W$ over a field $F$ is as follows: Let $v_1, ..., v_m$ be a basis for $V$ and let $w_1,...,w_n$ be a basis for $W$. We define $V \otimes W$ to be the set of formal linear combinations of the mn symbols $v_i \otimes w_j$. That is, a typical element of $V \otimes W$ is $$\sum c_{ij}(v_i \otimes w_j).$$ The space $V \otimes W$ is clearly a finite dimensional vector space of dimension mn. We define bilinear map $$B: V \times W \to V \otimes W$$ here is the formula $$B(\sum a_iv_i, \sum b_jw_j) = \sum_{i,j}a_ib_j(v_i \otimes w_j). $$

Why does $V \otimes W$ have to be a formal linear combinations of symbols $v_{i} \otimes w_j$, what would be wrong in defining $V \otimes W$ simply as a linear combination of symbols $v_i \otimes w_j$?


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    $\begingroup$ What exactly is the difference, in your mind, between a "formal linear combination" and a "linear combination"? $\endgroup$ – Jack M Nov 23 '16 at 18:02

The term "linear combination" has no meaning unless you are talking about elements of a vector space. The symbols $v_i\otimes w_j$ are no more than symbols, and there is not (yet) any vector space that they are elements of, so it is meaningless to talk about linear combinations of them. However, we can talk about "formal linear combinations", which means roughly "pretend they are elements of a vector space and write down symbols that would represent linear combinations of them".

More rigorously, you can say that a "formal linear combination" of the symbols $v_i\otimes w_j$ is just a function from the set of these symbols to $F$. Given such a function $f$, if $f(v_i\otimes w_j)=c_{ij}$, we represent $f$ by the notation $\sum c_{ij}v_i\otimes w_j$. The set of such functions then forms a vector space where addition and scalar multiplication are defined pointwise, and this addition and scalar multiplication corresponds to how you would expect expressions of the form $\sum c_{ij}v_i\otimes w_j$ to behave if they actually were linear combinations.


I propose you this answer: understanding the meaning of formal linear combination and tensor product

In particular: "Regarding the tensor product, we want it to be bilinear. In $R⟨V×W⟩$, it is never true that $(a+b)×c(a+b)×c$ is equal to $a×c+b×ca×c+b×c$. In fact, $0×0$ is not even the zero vector of $R⟨V×W⟩$"

  • $\begingroup$ Okay thanks. In the post you proposed, do you know why identifying every element $x \in S$ with the function that takes the value $1$ on $x$ and zero on other elements of $S$ allow any $F \in \mathbb{R}\langle s \rangle$ to be written uniquely as $F = \sum^{m}_{i=1}a_ix_i$? $\endgroup$ – Alex Nov 23 '16 at 18:24
  • $\begingroup$ Because $x \in S$ can be thought as a linear combination with just one element multiplied by 1, so it's a method to take into consideration only that point s, if you want a linear combination non trivial you have to consider different element of s etc... $\endgroup$ – Mattia Ghio Nov 23 '16 at 20:11
  • $\begingroup$ But how does he use this observation to go from defining a formal linear combination as a mapping $F: S \to \mathbb{R}$ to $F = \sum_{i}^{m}a_ix_i$ using the observation that any element $x \in S$ has a unique mapping from the $x$ to $1$ where the rest is equal to $0$? $\endgroup$ – Alex Nov 23 '16 at 21:43
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    $\begingroup$ $F(s)=\sum_{1}^{m}a_i x_i(s)$ if $s\neq x_i$ for all $i=1,...,m$ then $F(s)=0$ else $s=x_j$ and then $F(s)=a_j x_j(s=x_j)=a_j$ $\endgroup$ – Mattia Ghio Nov 24 '16 at 11:00
  • $\begingroup$ Thanks for all the assistance by the way. The answer in the recommened post does not address the question asked. He is asking why we require the quotient space $\mathbb{R}\langle V \times W \rangle / \mathcal{R} := V \otimes W$ (where $\mathcal{R} \subset \mathbb{R}\langle V \times W \rangle $ is an approproaitely chosen subspace of the free vector space) rather than just directly using the quotient space $(V \times W) / \mathcal{R}$. The answer just gives a reason why we need a quotient space of $\mathbb{R}(V \times W)$ rather than why we prefer this to a quotient space of $V \times W$. $\endgroup$ – Alex Nov 24 '16 at 16:56

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