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Suppose $V$ is an $n$-dimensional vector space over $\mathbb{R}$. Let $\mathcal{L}(V,V; \mathbb{R})$ denote the set of all bilinear maps $f: V\times V \rightarrow \mathbb{R}$, i.e., $f(v,w)$ is linear in each variable separately. (We can actually show that $\mathcal{L}(V,V; \mathbb{R})$ is a vector space over $\mathbb{R}$ under vector addition and scalar multiplication.)

I want to find a basis for $\mathcal{L}(V,V; \mathbb{R})$. I was given the following hint: If $e_1, \ldots, e_n$ is a basis for $V$ and $f \in \mathcal{L}(V,V; \mathbb{R})$ then the $f(e_i, e_j)$ play an important role.

I messed around with this hint:

Suppose $e_1, \ldots, e_n$ is a basis for $V$. Let $v,w \in V$ then we can express $v$ and $w$ as linear combinations of the $e_i$ vectors, $$ v=\sum_{i=1}^n a_i e_i$$ and $$ w=\sum_{k=1}^n b_k e_k.$$

By bilinearity of $f\in \mathcal{L}(V,V; \mathbb{R})$ we have $$f(v,w)= f\left( \sum_{i=1}^n a_i e_i, \sum_{k=1}^n b_k e_k \right)=\sum_{i=1}^n \sum_{k=1}^n a_ib_k f(e_i, e_k).$$

Here is where I get stuck: $f\in \mathcal{L}(V,V; \mathbb{R})$ is just one bilinear map in the space, but how can I find a set of bilinear maps in the space that spans this entire set of bilinear maps? Thank you!

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    $\begingroup$ You're almost there. Define the map $f_{i,k}:V\times V\to\mathbb{R}$ as follows. First, $f_{i,k}(e_m,e_n)=1$ if $m=i$, $n=k$ and 0 otherwise. Second, extend $f_{i,k}$ by (bi)linearity to a map on all of $V\times V$. Then check that these maps are indeed a basis. $\endgroup$ – symplectomorphic Aug 20 '16 at 4:51
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    $\begingroup$ A bilinear form is a covariant $2$-tensor on $V$. Let $(\varepsilon^j)$ be the dual basis for $V^*$. Then $\{\varepsilon^{i_1}\otimes\varepsilon^{i_2} : 1\leq i_1,i_2\leq n\}$ is a basis for $V^*\otimes V^*$. $\endgroup$ – Fei Li Aug 20 '16 at 4:55
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You have the idea right, that every bilinear form is defined (uniquely) by how it treats each pair of basis elements. Therefore, to make a basis for the space of bilinear forms, let $f_{i,j}(e_a, e_b) = 1$ iff $i=a$ and $j=b$, and $f_{i,j}(e_a, e_b)=0$ otherwise (and then these maps can be extended to all of $V\times V$ in the natural manner). Then $F=\{f_{i,j}: 1\leq i,j\leq n\}$ is a basis for the space of all bilinear forms on $V$. Based on your work, you should be able to represent any bilinear form as a linear combination of elements of $F$. Then you just need to show that the set is linearly independent, i.e. that no nontirival linear combination of elementsof $F$ can be a nonzero bilinear form (that is, every such form is nonzero for at least one input).

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    $\begingroup$ The maps $f_{i,j}$ as you've defined them are technically not bilinear forms on $V\times V$: you have defined them only on the set $B\times B$, where $B$ is the basis of $V$. You must extend them to $V$ by bilinearity. $\endgroup$ – symplectomorphic Aug 20 '16 at 4:53
  • $\begingroup$ I thought that was implied, but I edited. Thank you. $\endgroup$ – florence Aug 20 '16 at 4:57
  • $\begingroup$ @florence Ahh okay, I see. The space $\mathcal{L}(V\times V; \mathbb{R})$ is a dual space of $V\times V$? $\endgroup$ – Sarah Aug 20 '16 at 5:02
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    $\begingroup$ Yes. Keep in mind though, the dual space of $V\times V$ would be the set of all linear transformations from $V\times V$ to $\Bbb{R}$, but in general bilinear forms on $V$ aren't linear transformations with domain $V\times V$. Fei Li on the comments was taking the tensor product of basis elements of the dual space, which does give a set of bilinear forms. $\endgroup$ – florence Aug 20 '16 at 5:17

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