Let \(V\) be a finite-dimensional vector space over a subfield of the complex numbers and \(f\) a skew-symmetric bilinear form on \(V\). Show that \(f\) has rank 2 if and only if there exist linearly independent linear functionals \(L_1\), \(L_2\) on \(V\) such that \[ f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha). \]
Solution
1. (\(\Leftarrow\)) If \(f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha)\) with \(L_1, L_2\) linearly independent, then \(f\) has rank \(2\).
1.1. \(f\) is skew-symmetric.
\(f(\beta, \alpha) = L_1(\beta)L_2(\alpha) - L_1(\alpha)L_2(\beta) = -f(\alpha, \beta)\).
1.2. The matrix of \(f\) (in a basis extending a basis of \(\ker L_1 \cap \ker L_2\)) has rank \(2\).
\(f\) is the wedge \(L_1 \wedge L_2\); since \(L_1, L_2\) are linearly independent, the associated alternating form has rank \(2\) (its radical is \(\ker L_1 \cap \ker L_2\), of codimension \(2\)).
1.3. Hence \(f\) has rank \(2\).
2.
2. (\(\Rightarrow\)) If \(f\) has rank \(2\), then \(f = L_1 \wedge L_2\) for linearly independent \(L_1, L_2\).
2.1. A skew-symmetric form of rank \(2\) has a matrix (in a suitable basis) of the form \(\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}\) in a \(2\)-dimensional block, with zeros elsewhere.
the standard normal form of an alternating form: it is a direct sum of hyperbolic planes \(\begin{pmatrix}0&1\\-1&0\end{pmatrix}\) and a zero block; rank \(2\) means exactly one hyperbolic plane.
2.2. In this basis, \(f(\alpha, \beta) = \alpha_1 \beta_2 - \alpha_2 \beta_1\).
read off the matrix.
2.3. Define \(L_1(\alpha) = \alpha_1\) and \(L_2(\alpha) = \alpha_2\) (the coordinate functionals); these are linearly independent.
they are distinct coordinate functionals.
2.4. Then \(f(\alpha, \beta) = L_1(\alpha)L_2(\beta) - L_1(\beta)L_2(\alpha)\).
2 and 3.
3. Q.E.D.
1 and 2.