Let \(V\) be the space of all \(n \times 1\) matrices over \(C\) and \(f\) the bilinear form on \(V\) given by \(f(X, Y) = X^tY\). Let \(M\) belong to \(O(n, C)\). What is the matrix of \(f\) in the basis of \(V\) consisting of the columns \(M_1, M_2, \ldots, M_n\) of \(M\)?
Solution
Because \(M\in O(n,\mathbb C)\), it is invertible, so its columns \[ \mathfrak B=(M_1,\ldots,M_n) \] form a basis of \(V\).
The \((j,k)\) entry of the matrix of \(f\) in this basis is \[ f(M_j,M_k)=M_j^tM_k. \] Since \(M\) is orthogonal, \[ M^tM=I, \] and therefore \[ M_j^tM_k=\delta_{jk}. \] Hence \[ \boxed{[f]_{\mathfrak B}=I_n.} \]