Let \(F\) be a field, let \(V\) be the space of \(n\times1\) matrices over \(F\), and let \[ f(X,Y)=X^tAY \] be a non-degenerate symmetric bilinear form, where \(A\) is an invertible symmetric \(n\times n\) matrix. Let \(P\) be an invertible \(n\times n\) matrix whose columns form an ordered basis \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] of \(V\). Find the matrix whose columns are the vectors of the \(f\)-dual basis \[ \mathfrak B'=\{\alpha'_1,\ldots,\alpha'_n\}. \]
Solution
Let \[ Q= \begin{pmatrix} |&&|\\ \alpha'_1&\cdots&\alpha'_n\\ |&&| \end{pmatrix} \] be the unknown matrix whose columns form the \(f\)-dual basis.
1. Express the duality equations in matrix form.
The \((i,j)\) entry of \(P^tAQ\) is \[ \alpha_i^tA\alpha'_j =f(\alpha_i,\alpha'_j). \] By definition of the \(f\)-dual basis, this equals \(\delta_{ij}\). Therefore \[ P^tAQ=I. \]
2. Solve for the dual-basis matrix.
Both \(P\) and \(A\) are invertible, so the equation from step 1 gives \[ AQ=(P^t)^{-1} \] and hence \[ \boxed{Q=A^{-1}(P^t)^{-1}.} \]
Conversely, this matrix satisfies \[ P^tA\bigl(A^{-1}(P^t)^{-1}\bigr)=I, \] so its columns have exactly the required pairings with the columns of \(P\). Thus it is the unique matrix of the \(f\)-dual basis.