Let \(V\) be a finite-dimensional vector space and \(L_1, L_2\) linear functionals on \(V\). Show that the equation \[ f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha) \] defines a skew-symmetric bilinear form on \(V\). Show that \(f = 0\) if and only if \(L_1, L_2\) are linearly dependent.
Solution
1. \(f\) is bilinear.
\(L_1(\alpha)L_2(\beta)\) is bilinear in \((\alpha, \beta)\) (product of a linear functional in \(\alpha\) and one in \(\beta\)), and \(L_1(\beta)L_2(\alpha)\) is bilinear too; their difference is bilinear.
2. \(f\) is skew-symmetric.
\(f(\beta, \alpha) = L_1(\beta)L_2(\alpha) - L_1(\alpha)L_2(\beta) = -f(\alpha, \beta)\).
3. (\(\Leftarrow\)) If \(L_1, L_2\) are linearly dependent, then \(f = 0\).
3.1. If \(L_2 = cL_1\) (or \(L_1 = cL_2\)), then \(f(\alpha, \beta) = L_1(\alpha)(cL_1(\beta)) - L_1(\beta)(cL_1(\alpha)) = cL_1(\alpha)L_1(\beta) - cL_1(\beta)L_1(\alpha) = 0\).
substitute the linear dependence.
4. (\(\Rightarrow\)) If \(f = 0\), then \(L_1, L_2\) are linearly dependent.
4.1. Suppose \(L_1, L_2\) are linearly independent.
assume the contrary.
4.2. Then there exist \(\alpha, \beta \in V\) with \(L_1(\alpha) = 1\), \(L_2(\alpha) = 0\), \(L_1(\beta) = 0\), \(L_2(\beta) = 1\).
since \(L_1, L_2\) are independent, the map \(V \to k^2\), \(v \mapsto (L_1(v), L_2(v))\), is surjective, so we can find \(\alpha\) mapping to \((1,0)\) and \(\beta\) mapping to \((0,1)\).
4.3. Then \(f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha) = 1 \cdot 1 - 0 \cdot 0 = 1 \neq 0\).
2.
4.4. This contradicts \(f = 0\).
3.
4.5. Hence \(L_1, L_2\) are linearly dependent.
<2>1–<2>4.
5. Q.E.D.
1, 2, 3, and 4.