Let \(F\) be a field and let \(V\) be the space of \(n\times1\) matrices over \(F\). Suppose \(A\) is a fixed \(n\times n\) matrix over \(F\) and \[ f(X,Y)=X^tAY \] is a symmetric, non-degenerate bilinear form on \(V\). Let \(B\) be an \(n\times n\) matrix over \(F\), and let \(T\) be the linear operator \[ T(X)=BX. \] Find the adjoint \(T'\) relative to \(f\) from Exercise 11.
Solution
Because \(f\) is symmetric, \[ A^t=A, \] and because \(f\) is non-degenerate, \(A\) is invertible.
1. Translate the adjoint identity into a matrix equation.
Suppose \(T'\) has matrix \(C\), so \(T'(Y)=CY\). Then \[ \begin{aligned} f(TX,Y) &=(BX)^tAY\\ &=X^tB^tAY, \end{aligned} \] whereas \[ f(X,T'Y)=X^tACY. \] Thus \[ f(TX,Y)=f(X,T'Y) \] for every \(X,Y\) if and only if \[ B^tA=AC. \]
2. Solve for the adjoint matrix.
Since \(A\) is invertible, \[ C=A^{-1}B^tA. \] Therefore \[ \boxed{T'(Y)=A^{-1}B^tAY.} \]
Direct substitution verifies the formula: \[ f(X,T'Y) =X^tAA^{-1}B^tAY =X^tB^tAY =f(TX,Y). \] By the uniqueness proved in Exercise 11, this is the relative adjoint.