Let \(V\) be a finite-dimensional vector space and let \(f\) be a non-degenerate symmetric bilinear form on \(V\). Associated with \(f\) is the natural isomorphism \[ L_f:V\longrightarrow V^*, \qquad L_f(\alpha)(\beta)=f(\alpha,\beta). \] Using \(L_f\), show that for each basis \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] of \(V\) there exists a unique basis \[ \mathfrak B'=\{\alpha'_1,\ldots,\alpha'_n\} \] such that \[ f(\alpha_i,\alpha'_j)=\delta_{ij}. \] Then show that every \(\alpha\in V\) satisfies \[ \alpha=\sum_{i=1}^n f(\alpha,\alpha'_i)\alpha_i =\sum_{i=1}^n f(\alpha_i,\alpha)\alpha'_i. \]
Solution
Let \[ \mathfrak B^*=\{\lambda_1,\ldots,\lambda_n\} \] be the ordinary dual basis in \(V^*\), so \[ \lambda_j(\alpha_i)=\delta_{ij}. \]
1. Construct the unique \(f\)-dual basis.
Since \(L_f\) is an isomorphism, for each \(j\) there is a unique vector \(\alpha'_j\) such that \[ L_f(\alpha'_j)=\lambda_j. \] By symmetry of \(f\), \[ f(\alpha_i,\alpha'_j) =f(\alpha'_j,\alpha_i) =L_f(\alpha'_j)(\alpha_i) =\lambda_j(\alpha_i) =\delta_{ij}. \]
The vectors \(\alpha'_1,\ldots,\alpha'_n\) form a basis because their images under the isomorphism \(L_f\) form the basis \(\mathfrak B^*\) of \(V^*\).
If another basis \((\beta'_1,\ldots,\beta'_n)\) had the same pairings, then for each \(j\) the two functionals \(L_f(\beta'_j)\) and \(L_f(\alpha'_j)\) agree on every basis vector \(\alpha_i\), hence are equal. Injectivity of \(L_f\) gives \(\beta'_j=\alpha'_j\). Thus the \(f\)-dual basis is unique.
2. Expand a vector in the original basis using the \(f\)-dual basis.
Write \[ \alpha=\sum_i c_i\alpha_i. \] Pairing with \(\alpha'_j\) gives \[ f(\alpha,\alpha'_j) =\sum_i c_i f(\alpha_i,\alpha'_j) =c_j. \] Therefore \[ \boxed{ \alpha=\sum_{i=1}^n f(\alpha,\alpha'_i)\alpha_i.} \]
3. Expand a vector in the \(f\)-dual basis.
Write \[ \alpha=\sum_i d_i\alpha'_i. \] Pairing on the left with \(\alpha_j\) gives \[ f(\alpha_j,\alpha) =\sum_i d_i f(\alpha_j,\alpha'_i) =d_j. \] Hence \[ \boxed{ \alpha=\sum_{i=1}^n f(\alpha_i,\alpha)\alpha'_i.} \]