Let $B$ be a non-empty set. Is it possible to construct a vector space $X$ whose the elements of its base are the elements of the set $B$?

47 Views Asked by At

I'm trying to understand the definition of tensor product between two vector spaces, but in some demonstrations of the existence of the tensor product there is the following assumption: "Let $U$ and $V$ be vector spaces over a field $K$ and $\langle U\times V\rangle $ be a vector space over the same field whose base is $U\times V$". What I don't understand is why there's a vector space $\langle U\times V\rangle $ whose base is the set $U\times V$. By the way some write this assumption it seems to me that given a non-empty set $B$ is possible to construct a vector space $X$ whose base is the set $B$.

Something I found that may be related to my question is about free vector space which is described in the following link: https://math.berkeley.edu/~shiyu/s15capstone/materials/Capstone_Course%20(7).pdf

But, given a vector space $V$ over a field $K$, I don't know how to construct a free vector space whose base is the set $V$.

Please explain as detailed as possible how to construct a vector space whose base is a given set. I would like an explanation to finally understand the construction of a tensor product using this idea of constructing an appropriate vector space.

1

There are 1 best solutions below

1
On BEST ANSWER

Given some set $B$, consider the set $F_B$ of all functions $f\colon B\to K$ such that $f(b)\not=0$ for at most finitly many $b\in B$. This is a vector space over $K$, in fact a subspace of the space $K^B$ of all mappings defined on $B$ with values in $K$. A basis of $F_B$ is given by the mappings $c_b, b\in B$, where $c_b(b'):=0$ for $b'\not=b$ and $c_b(b):=1$. Finally one may identify $b$ with $c_b$.