Let \(V\) be a finite-dimensional complex vector space and let \(f\) be a non-degenerate symmetric bilinear form on \(V\). Show that there is a basis \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] whose \(f\)-dual basis is \(\mathfrak B\) itself.
Solution
1. Choose a basis in which the symmetric form is diagonal.
By diagonalization of symmetric bilinear forms, there is a basis \[ e_1,\ldots,e_n \] such that \[ f(e_i,e_j)=0 \qquad(i\ne j). \] Write \[ \lambda_i=f(e_i,e_i). \]
Every \(\lambda_i\) is nonzero. Indeed, if \(\lambda_i=0\), then \(e_i\) is orthogonal to every basis vector \(e_j\), hence to all of \(V\). This would put \(e_i\) in the radical of \(f\), contradicting non-degeneracy.
2. Rescale the diagonal basis to make every diagonal value equal to one.
Because the scalar field is \(\mathbb C\) and \(\lambda_i\ne0\), choose \(\mu_i\in\mathbb C^\times\) such that \[ \mu_i^2=\lambda_i. \] Set \[ \alpha_i=\mu_i^{-1}e_i. \] Then for \(i\ne j\), \[ f(\alpha_i,\alpha_j)=0, \] while \[ f(\alpha_i,\alpha_i) =\mu_i^{-2}f(e_i,e_i) =\mu_i^{-2}\lambda_i =1. \] Thus \[ \boxed{f(\alpha_i,\alpha_j)=\delta_{ij}.} \]
3. Identify the \(f\)-dual basis.
By definition, the \(f\)-dual basis to \[ \mathfrak B=\{\alpha_1,\ldots,\alpha_n\} \] is the unique basis \(\{\alpha'_1,\ldots,\alpha'_n\}\) satisfying \[ f(\alpha_i,\alpha'_j)=\delta_{ij}. \] Step 2 shows that the vectors \(\alpha'_j=\alpha_j\) already satisfy these relations. Uniqueness of the \(f\)-dual basis from Exercise 13 therefore gives \[ \boxed{\mathfrak B'=\mathfrak B.} \]