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!
Best Answer
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).