Show that for a bilinear skew-symmetric function, mapping $\mathbb R^n \times \mathbb R^n \to \mathbb R$,
$b(x,y)=\langle Bx,y\rangle$, where $\langle x,y\rangle$ is the inner product $\sum_j x_jy_j$.
My work:
Define the matrix B with entries $B_{ij}$ = $b(e_i,e_j)$
Using the standard basis for $R^n$, write $\vec x = \sum_i x_ie_i$ and $\vec y=\sum_j y_je_j.$
Then we have that
$$b(\vec x,\vec y) = b\big(\sum_i x_ie_i,\sum_j y_je_j\big)$$ $$= \big(\sum_i \sum_jb(x_ie_i, y_je_j)\big)$$ $$= \big(\sum_i \sum_jx_iy_jb(e_i, e_j)\big)$$ $$= \big(\sum_i \sum_jx_iy_jB_{ij}\big)$$ $$= \big(\sum_ix_i \sum_jy_jB_{ij}\big)$$ $$= \big(\sum_ix_i \sum_jB_{ij}y_j\big)$$ $$= \big(\sum_ix_i \langle B_i,\vec y\rangle \big)$$ $$= \big(\sum_i \langle B_ix_i,\vec y\rangle \big)$$ $$= \langle B\vec x,\vec y\rangle,$$ as required.
I am not so sure about the last 3 equalities, going from rows $B_i$ to the whole matrix $B$ seems a little awkward -- but the logic seems fine to me, using just bilinearity and the inner product definition.
Any comments or suggestions are welcome.
Thanks,