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I have a question about understanding the meaning of formal linear combination.

Let S be a set, the free vector space $\mathbb{R}\langle s\rangle$ on S is defined as the set of all formal linear combination of elements of S with real coefficients. The formal linear combination is a function F: S$\to$$\mathbb{R}$ such that F(s)=0 for all but finitely many s$\in$S. Identifying every element x$\in$S with the function that takes the value 1 on x and zero on other elements of S, any element F$\in$$\mathbb{R}\langle s\rangle$ can be written uniquely in the form F=$\sum_{i=1}^{m}a_ix_i$, where $x_i$ are the elements of S for which F($x_i$)$\neq$0 and $a_i$=F($x_i$).

Here is first question: why we define things like this? why we just say the finite sum of elements in S?

My second question is: Since the free vector space can be used to define tensor product of real vector spaces V and W. Then why we use $\mathbb{R}\langle V\times W\rangle$ modding out some equivalence class instead of use the finite sum of elements in the form $v_i\times w_i$ where $v_i\in V$ and $w_i\in W$ modding out that equivalence class? Thanks for any hint!

user146507
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1 Answers1

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Sum of things in $S$? $S$ is just a set; what can the sum of things in $S$ mean?

Once you've constructed $\mathbf{R}\langle s \rangle$, it makes sense to talk about the sum of finitely many things in $S$ using the $+$ operation of $\mathbf{R}\langle s \rangle$ for addition. But without having chosen a $+$ operation, talking about sums is nonsensical.

(also, it's important that we want linear combinations, not merely finite sums)

Regarding the tensor product, we want it to be bilinear. In $\mathbf{R}\langle V \times W \rangle$, it is never true that $(a+b) \times c$ is equal to $a \times c + b \times c$. In fact, $0 \times 0$ is not even the zero vector of $\mathbf{R}\langle V \times W \rangle$!

  • Thanks, I think I get what you say about the first question. As for the second one, we mod out the equivalence class to make it bilinear. V$\times$W is a vector space, I guess we just follow the way we deal with set to define the free vector space. Is that right? – user146507 Nov 19 '14 at 23:21
  • Right. A useful notational device is, for $s \in S$, to let $[s] \in \mathbf{R}[S]$ denote the corresponding vector (you might use angle brackets rather than square brackets, based on your notation). In the case that $S$ is a vector space itself, this device lets us easily distinguish between $a+b \in S$, $[a+b] \in \mathbf{R}[S]$, and $[a] + [b] \in \mathbf{R}[S]$. Or similarly if $r$ is a scalar, we can easily tell apart $r[a]$ and $[ra]$. –  Nov 19 '14 at 23:34
  • Thank you very much! I think I see the intuition. – user146507 Nov 20 '14 at 00:51
  • @Hurkyl Hi could you maybe explain how formal linear combination which is given as mapping $F: S \to \mathbb{R}$ becomes $F = \sum_{i}^{m}a_ix_i$ using the observation that any element $x \in S$ has a unique mapping from the $x$ to $1$ where the rest is equal to $0$? Thanks. – Alex Nov 23 '16 at 21:40