Find all bilinear forms on the space of \(n \times 1\) matrices over \(\RR\) which are invariant under \(O(n, R)\).
Solution
Write \[ f(x,y)=x^tBy \] for a unique matrix \(B\in M_n(\mathbb R)\).
1. Orthogonal invariance is equivalent to commuting with every orthogonal matrix.
For \(A\in O(n,\mathbb R)\), \[ f(Ax,Ay)=x^tA^tBAy. \] Thus invariance for all \(x,y\) is equivalent to \[ A^tBA=B. \] Since \(A^{-1}=A^t\), this is equivalent to \[ BA=AB. \]
2. Commuting with all coordinate sign changes forces \(B\) to be diagonal.
For each \(k\), let \[ D_k=\operatorname{diag}(1,\ldots,1,-1,1,\ldots,1), \] with \(-1\) in the \(k\)th position. Then \(D_k\in O(n,\mathbb R)\).
If \(B=(b_{ij})\) and \(i\ne j\), use \(D_iB=BD_i\). Its \((i,j)\) entry gives \[ -b_{ij}=b_{ij}, \] so \(b_{ij}=0\). Hence \(B\) is diagonal.
3. Commuting with coordinate permutations forces a diagonal \(B\) to be scalar.
Write \[ B=\operatorname{diag}(b_1,\ldots,b_n). \] Every permutation matrix is orthogonal. If \(P_{ij}\) exchanges coordinates \(i\) and \(j\), then \[ BP_{ij}=P_{ij}B \] forces \(b_i=b_j\). Hence all diagonal entries are equal and \[ B=cI \] for some \(c\in\mathbb R\).
4. Conclude the classification.