Describe explicitly all bilinear forms \(f\) on \(\RR^3\) with the property that \(f(\alpha, \beta) = f(\beta, \alpha)\) for all \(\alpha, \beta\).
Solution
Let \[ \alpha=(x_1,x_2,x_3), \qquad \beta=(y_1,y_2,y_3). \]
1. A bilinear form is determined by its matrix in the standard basis.
Let \(e_1,e_2,e_3\) be the standard basis and put \[ a_{ij}=f(e_i,e_j). \] By bilinearity, \[ f(\alpha,\beta) =\sum_{i,j=1}^3 a_{ij}x_i y_j. \] Thus every bilinear form is represented uniquely by the matrix \(A=(a_{ij})\) through \[ f(\alpha,\beta)=\alpha^tA\beta. \]
2. Symmetry of the form is exactly symmetry of its matrix.
If \[ f(\alpha,\beta)=f(\beta,\alpha) \] for all \(\alpha,\beta\), then in particular \[ a_{ij}=f(e_i,e_j)=f(e_j,e_i)=a_{ji}. \] Hence \(A=A^t\).
Conversely, if \(A=A^t\), then \[ f(\beta,\alpha) =\beta^tA\alpha =\alpha^tA^t\beta =\alpha^tA\beta =f(\alpha,\beta). \] So the form is symmetric.
3. Write all such forms explicitly.
An arbitrary symmetric \(3\times3\) real matrix has the form \[ A= \begin{pmatrix} a&d&e\\ d&b&g\\ e&g&c \end{pmatrix}, \] with arbitrary real parameters \(a,b,c,d,e,g\). Therefore the symmetric bilinear forms on \(\mathbb R^3\) are exactly \[ \boxed{ \begin{aligned} f(\alpha,\beta) ={}&a x_1y_1+b x_2y_2+c x_3y_3\\ &+d(x_1y_2+x_2y_1) +e(x_1y_3+x_3y_1)\\ &+g(x_2y_3+x_3y_2), \end{aligned}} \] where \(a,b,c,d,e,g\in\mathbb R\) are arbitrary. Conversely every expression of this form is visibly bilinear and symmetric.