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I was hoping someone could help me with the above question (give an isomorphism between the linear spaces $U\times V$ and $L(U,V)$. I have a hunch that I should work with the bases of U and V but the exact method is unclear to me. I just need to see it worked out completely once so I can proceed with the rest of my questions. Thank you.

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What if $U=V=\mathbb{R}$ for instance? –  1015 Feb 14 '13 at 14:57
3  
I guess you mean $\otimes$ rather than $\times$ (since otherwise it is not true). Also, I suppose these are finite dimensional spaces? –  Tobias Kildetoft Feb 14 '13 at 14:57
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A good thing is to introduce the intermediate space $U^* \otimes V$. –  Damien L Feb 14 '13 at 14:58

1 Answer 1

Assume that $U$ is finite dimensional. Also, as the comments say, it is about rather the tensor product $U\otimes V$ of the vector spaces, and not the cartesian one.

Fix a basis $u_1,..,u_n$ in $U$. Then $L(U,V)\cong V^n$ as the linear maps $f:U\to V$ are uniquely determined by the arbitrary values $f(u_1),..,f(u_n)\in V$.

On the other hand, $U\otimes V$ consists of finite (formal) sums $\sum_k x_k\otimes y_k$ where $x_k\in U,\, y_k\in V$. Now write each $x_k$ in the given basis, so it will get the form $\sum_{k=1}^n u_k\otimes v_k$ for some $v_k\in V$. You need that this latter sum is $0$ iff all $v_k=0$, that is, the elements of $U\otimes V$ can uniquely be expressed as $\sum_{k=1}^n u_k\otimes v_k$ for arbitrary $v_k\in V$'s. So, again it yields $U\otimes V\cong V^n$.

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