Let \(f\) be a non-degenerate form (Exercises 6 and 7) on a finite-dimensional space \(V\). Let \(L\) be a linear functional on \(V\). Show that there exists one and only one vector \(\beta\) in \(V\) such that \(L(\alpha) = f(\alpha, \beta)\) for all \(\alpha\).
Solution
Choose an ordered basis of \(V\). Let \(A\) be the matrix of \(f\) in this basis, so for coordinate columns \(X,Y\), \[ f(X,Y)=Y^*AX. \] Since \(f\) is non-degenerate, Exercise 7 shows it is both left and right non-degenerate. Equivalently, the square matrix \(A\) is invertible.
Write the linear functional \(L\) in coordinates as \[ L(X)=\ell^tX \] for a unique coefficient column \(\ell\).
1. Existence.
Set the row vector \[ r=\ell^tA^{-1}. \] There is a unique column \(Y\) such that \[ Y^*=r. \] Let \(\beta\) be the vector with coordinate column \(Y\). Then for every \(\alpha\) with coordinate column \(X\), \[ \begin{aligned} f(\alpha,\beta) &=Y^*AX\\ &=rAX\\ &=\ell^tA^{-1}AX\\ &=\ell^tX\\ &=L(\alpha). \end{aligned} \] Thus such a representing vector exists.
2. Uniqueness.
Suppose \(\beta_1\) and \(\beta_2\) both represent \(L\). Then for every \(\alpha\), \[ f(\alpha,\beta_1-\beta_2)=0. \] By right non-degeneracy of \(f\), \[ \beta_1-\beta_2=0. \] Hence \(\beta_1=\beta_2\).
Therefore every linear functional has a unique representation \[ \boxed{L(\alpha)=f(\alpha,\beta)} \] through the non-degenerate form \(f\).