# Representing a Bilinear Form in a Matrix

Let $$b : \mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}$$ be a bilinear form, and $$\langle,\rangle$$ be the standard inner product of the Euclidean space and $$e_1,...e_n$$ be the standard basis for the Euclidean space.

Then, for arbitrary $$x$$ and $$y$$ in $$\mathbb{R}^n$$, I want to construct a matrix $$B$$ such that $$b(x,y)=\langle Bx,y\rangle$$.

Here, is it correct that the matrix $$B$$ is defined to be the 'transpose' of matrix $$B'$$ whose component in the $$i$$th row and $$j$$th column is $$b(e_i,e_j)$$?

Because $$B$$ must be in the first argument of the inner product, I think it is necessary to take the transpose. What do you think?

Yes because note that the standard inner product in $$\Bbb R^n$$ can be expressed like so: $$\langle p, q \rangle = p^Tq$$ where $$p^T$$ is the transpose of $$p$$.
Hence $$b(x, y) = \langle Bx, y \rangle = (Bx)^Ty = x^TB^Ty$$.
So $$b(e_i, e_j) = e_i^T B^T e_j$$ and the operation $$e_i^T B^T e_j$$ selects the $$i^\text{th}$$ row of $$B^T$$ and the $$j^\text{th}$$ row of $$B^T$$.
Or in other words $$b(e_i, e_j) = [B^T]_{i, j} = B_{j, i}$$. In your notation the matrix $$B^T$$ is of course $$B'$$.