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Consider the set of all bilinear forms $\text{Bil}_F(X \times Y)$ over finite dimensional vector spaces $X$ and $Y$ with common ground field F. If we choose bases for $(u_1, \dots, u_m)$ for $X$ and $(v_1, \dots , v_m)$ for Y, a basis for the tensor product of $X$ and $Y$ is given by $\{u^i \otimes v^j | i = 1, \dots, m; j=1, \dots n \}$ where $u^i$ and $v^i$ denote the respective dual bases. Now, it is a theorem that the tensor product $X \bigotimes Y$ is isomorphic to $\text{Bil}_F(X \times Y)$ and counting the $m \cdot n$ basis elements of $X \bigotimes Y$ we know that the tensor product is isomorphic to the set of all $m \times n$ matrices over $F$, $\mathcal{M}^m_n(F)$ which also has $m \cdot n$ basis elements and therefore, $\text{Bil}_F(X \times Y) \approx \mathcal{M}^m_n(F)$


Now, my question is: Assuming the above argument makes sense, this seems like a lot of technology to prove this isomorphism. Is there a better/more direct way to do this that doesn't involve, for example, constructions such as the tensor product?


I know that there is a unique matrix associated with given bilinear form that has entries determined by $B_{ij} = B(u_i, v_j)$ and this insures that one can, in principle, construct at least an injection between $\text{Bil}_F(X \times Y)$ and $\mathcal{M}^m_n(F)$. Also, given any matrix $A$, the mapping $B(x,y) = x^T A y$ is a bilinear form. It seems that a good candidate for an isomorphism might be $$ \phi(A)(x,y) = x^T A y $$ but it's not clear to me how this is necessarily a bijection.

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If you pick a basis of both $X$ and $Y$, there's an obvious induced basis of the space of bilinear forms (and this is already enough to prove the (noncanonical) isomorphism you want, since this basis has size $(\dim X)(\dim Y)$). Can you show that each element of this basis can be realized by a bilinear form of the form $x^T A y$? – Qiaochu Yuan Feb 8 '12 at 18:37
Well, the invocation of the tensor product is not entirely correct here anyway. The space of bilinear maps $X \times Y \to F$ is (naturally) isomorphic to $X^* \otimes Y^*$, not $X \otimes Y$. This is important when $X$ or $Y$ are infinite-dimensional. – Zhen Lin Feb 8 '12 at 18:43

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