Find all skew-symmetric bilinear forms on \(\RR^3\).
Solution
1. A skew-symmetric bilinear form \(f\) on \(\RR^3\) has matrix \(A = (a_{ij})\) with \(a_{ij} = f(e_i, e_j)\) and \(A^t = -A\).
skew-symmetry means \(a_{ij} = -a_{ji}\), so \(a_{ii} = 0\).
2. Hence \(A\) has the form\[A = \begin{pmatrix} 0 & a & b \\ -a & 0 & c \\ -b & -c & 0 \end{pmatrix}\]for arbitrary real \(a, b, c\).
1 (the diagonal is zero and the upper triangle determines the lower triangle).
3. Therefore every skew-symmetric form on \(\RR^3\) is\[f(x, y) = a(x_1 y_2 - x_2 y_1) + b(x_1 y_3 - x_3 y_1) + c(x_2 y_3 - x_3 y_2)\]for some \(a, b, c \in \RR\).
2, expanding \(x^t A y\).
4. Equivalently, \(f(x, y) = \det\begin{pmatrix} a & b & c \\ x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \end{pmatrix}\)-type expression: \(f(x,y) = (a, b, c) \cdot (x \times y)\), the dot product of the vector \((a,b,c)\) with the cross product \(x \times y\).
3, recognizing the cross product components.
5. Hence the space of skew-symmetric forms on \(\RR^3\) is \(3\)-dimensional, parametrized by \((a, b, c) \in \RR^3\).
3.
6. Q.E.D.
5.