Let \(V\) be a finite-dimensional vector space with a non-degenerate symmetric bilinear form \(f\). Let \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] be a basis, let \[ \mathfrak B'=\{\alpha'_1,\ldots,\alpha'_n\} \] be its \(f\)-dual basis, and let \[ A=[f]_{\mathfrak B}. \] Show that \[ \alpha'_j=\sum_{i=1}^n (A^{-1})_{ij}\alpha_i. \]
Solution
Write \[ \alpha'_j=\sum_{k=1}^n c_{kj}\alpha_k, \] and let \[ C=(c_{kj}) \] be the matrix whose \(j\)th column is the coordinate vector of \(\alpha'_j\) in \(\mathfrak B\).
1. Translate the dual-basis conditions into a matrix equation.
By definition of the Gram matrix, \[ A_{ik}=f(\alpha_i,\alpha_k). \] Therefore \[ \begin{aligned} f(\alpha_i,\alpha'_j) &=f\left(\alpha_i,\sum_k c_{kj}\alpha_k\right)\\ &=\sum_k A_{ik}c_{kj}\\ &=(AC)_{ij}. \end{aligned} \] The \(f\)-dual-basis relation says \[ f(\alpha_i,\alpha'_j)=\delta_{ij}, \] so \[ AC=I. \]
2. Solve for the coordinate matrix of the \(f\)-dual basis.
Non-degeneracy of \(f\) implies that its Gram matrix \(A\) is invertible. Hence \[ C=A^{-1}. \] Thus the \(j\)th column gives \[ \boxed{ \alpha'_j=\sum_{i=1}^n(A^{-1})_{ij}\alpha_i.} \]
Since \(A\) is symmetric, \(A^{-1}\) is symmetric as well, so equivalently the coefficient may be written \((A^{-1})_{ji}\).