$\{u_{i} \otimes w_{j} \}_{i , j}$ forms a basis for $U \otimes W$

Question

Suppose $U$ and $W$ are $k$-vector spaces with bases $\{u_{i}\}_{i=1}^{n}$ and $\{w_{j}\}_{j=1}^{m}$.

How to prove that $\{u_{i} \otimes w_{j} \}_{i , j}$ forms a basis for $U \otimes W$ ?

Answer 1

Hint: first prove that they span $U \otimes W$. Now for linear independence construct a bilinear map $B=U \times W \to k$ such that $B(u_s,w_l)$ is not zero only for $s=i$ and $l=j$.

Answer 2

Let $\;u\in U\;,\;w\in W\implies\;$ there exist linear combinations

$$u=\sum_{i=1}^na_iu_i\;,\;\;\sum_{j=1}^m b_jw_j\implies$$

$$v\otimes w=\sum_{i=1}^na_iu_i\otimes\sum_{j=1}^m b_jw_j=\left(\sum_{i=1}^na_i\sum_{j=1}^mb_j\right)u_i\otimes w_j$$

For linear independence we could try a dual basis of some basis of the free vector space $\;U\times W\;$ over $\;k\;$. Have you tried this?

Answer 3

-I'm not sure whether you mean what I understand from your symbols but here's what I think-

You need to prove 1. linear independence and 2. spanning property. For 1. take a linear combination equal to 0 ( eg λ1(u1,0) + ... λn(un,0) + μ1(0,w1)+...+μm(0,wm) = 0 ) and then use the linear independence of u and w (since they're bases) to get that λi=0 for i=1,..n and μi=0 for i=0,..,m. for 2. suppose (a,b) is an element of UxW and then write a and b as elements of U and W respectively,using the spanning property of their bases. With little work you'd have (a,b)=(λ1*u1,0)+...+(λν*un,0) + (0,μ1*w1)+...(0,μm*wm)=0 => λ1(u1,0) + ... λν(un,0) + μ1(0,w1)+..μm(0,wm) which means you can write a random element of UxW as a linear combination of ui x wi so you're done.

Answer 4

There is a book of Do Carmo, Differential forms and applications. There you will find an excellent explanation of this topic.