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My reading (link provided for completeness only, clicking is not necessary) defines the tensor product space as follows:

Let $V$ and $W$ be vector spaces. The symbol $v\otimes w$ is defined to be the bilinear formal product satisfying: for each $v_1,v_2\in V$ and $w_1,w_2\in W$, \begin{align*} (\alpha_1v_1+\alpha_2v_2)\otimes w&=\alpha_1v_1\otimes w+\alpha_2v_2\otimes w,\\ v\otimes(\alpha_1w_1+\alpha_2w_2)&=\alpha_1v\otimes w_1+\alpha_2v\otimes w_2, \end{align*} for all scalars $\alpha_1,\alpha_2$. Then $V\otimes W$ is the vector space of all linear combinations $\sum\alpha_iv_i\otimes w_i$ for $v_i\in V$ and $w_i\in W$.

Question: shortly after the definition above, the author comments that $V\otimes W$ is the dual of the space of bilinear functionals on $V\times W$. Why is this please? If possible, could you help with me with a gentle example?

Attempt: Let $A$ be the space of bilinear functionals on $V\times W$. Then, each $\alpha\in A$ is a bilinear functionals on $V\times W$ and the dual space $A^*$ is the space of linear functionals on $A$. But I'm confused as to why $A^*$ can be interpreted as $V\otimes W$.

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There is a natural bijection between linear maps $f: V \otimes W \to X$, and bilinear maps $g: V \times W \to X$, where $X$ is any vector space.

(Going one way, you define $g(v, w) = f(v \otimes w)$, and going to other you define $f(v \otimes w) = g(v, w)$, which is well-defined because $g$ is bilinear.)

In particular, if you take $X = k$, then there is a bijection between linear maps $V \otimes W \to k$ and bilinear maps $V \times W \to k$. Taking the dual of both sides, and recalling that (for finite dimensional $V$, $W$) the double dual of a space is canonically isomorphic to the original space, we have that $V \otimes W$ is isomorphic to the dual space of the space of all bilinear maps $V \times W \to k$, i.e. the dual of the space of all bilinear functionals.

Edit: more concretely: if we have an element $v \otimes w \in V \times W$, we then define an element of $A^*$ by taking $a \in A$ to $a(v, w)$. This is well-defined because $a$ is bilinear. The fact that this is an isomorphism is trickier, and it's probably easier to go via the above. It probably comes out reasonably nicely if you pick bases for $V$ and $W$, however.

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