A soft question on linear algebra Let $f$ be any skew-symmetric bilinear form on $\Bbb R^3$. I was trying to prove that there are linear functionals $L_{1},L_{2}$ such that $$f(\alpha,\beta) = L_{1}(\alpha)L_{2}(\beta) - L_{1}(\beta)L_{2}(\alpha)$$
I only know the definitions of the terms in the above problem.Please provide an answer to the problem...
 A: Hints:


*

*Represent $f$ with respect to the standard basis by a matrix $A$ so that $f(v,w) = v^T A w$ and show that $A^T = -A$. Since $n = 3$, show that this implies that $\det(A) = 0$.

*Choose $v_0$ such that $Av_0 = 0$ and complete $v_0$ to a basis $(v_0,v_1,v_2)$ of $\mathbb{R}^3$. With respect to this basis, $f$ will be represented by a matrix of the form
$$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & a \\ 0 & -a & 0 \end{pmatrix}. $$
Denote by $(\phi_0,\phi_1,\phi_2)$ the corresponding dual basis.

*Take $L_1 = a \phi_1$ and $L_2 = \phi_2$ and check that this works.

A: In $\mathbb{R}^3$, there is a non zero vector $u$ such that $L(u, \beta) = 0$ for all $\beta$. Choose 2 vectors $v, w$ such that $(u, v, w)$ is a basis of $\mathbb{R}^3$. For any $x\in \mathbb{R}^3$, we can decompose $x$ on this basis:
$$x = c_u(x) u + c_v(x)v + c_w(x)w$$
The component applications $c_u, c_v, c_w$ are linear forms. Now let us compute
$$L(\alpha, \beta)= c_v(\alpha)c_w(\beta) L(v, w) + c_w(\alpha) c_v(\beta) L(w, v)$$
We have $L(w,v) = -L(v, w)$, so if we define
$$L_1(\alpha) = c_v(\alpha) L(v, w)\quad L_2(\alpha) = c_w(\alpha)$$
we obtain the result.
