Describe the bilinear forms on \(\RR^3\) which satisfy \(f(\alpha, \beta) = -f(\beta, \alpha)\) for all \(\alpha, \beta\).
Solution
Let \[ \alpha=(x_1,x_2,x_3), \qquad \beta=(y_1,y_2,y_3). \]
1. The matrix of such a form must be skew-symmetric.
Write \[ f(\alpha,\beta)=\alpha^tA\beta, \qquad A=(a_{ij}), \] where \[ a_{ij}=f(e_i,e_j). \] The hypothesis gives \[ a_{ij}=f(e_i,e_j)=-f(e_j,e_i)=-a_{ji}. \] In particular, taking \(i=j\) yields \[ a_{ii}=-a_{ii}, \] and since the scalar field is \(\mathbb R\), \[ a_{ii}=0. \] Thus \[ A^t=-A. \]
2. Conversely, every skew-symmetric matrix defines such a form.
If \(A^t=-A\), then \[ f(\beta,\alpha) =\beta^tA\alpha =\alpha^tA^t\beta =-\alpha^tA\beta =-f(\alpha,\beta). \] Hence skew-symmetry of the matrix is equivalent to the required identity.
3. Write all such forms explicitly.
An arbitrary real skew-symmetric \(3\times3\) matrix is \[ A= \begin{pmatrix} 0&a&b\\ -a&0&c\\ -b&-c&0 \end{pmatrix}, \] with \(a,b,c\in\mathbb R\). Therefore the desired bilinear forms are exactly \[ \boxed{ \begin{aligned} f(\alpha,\beta) ={}&a(x_1y_2-x_2y_1)\\ &+b(x_1y_3-x_3y_1)\\ &+c(x_2y_3-x_3y_2), \end{aligned}} \] where \(a,b,c\) are arbitrary real scalars.