Let \(V\) be the space of \(n \times 1\) matrices over \(C\), with the inner product \((X \mid Y) = Y^*GX\) (where \(G\) is an \(n \times n\) matrix such that this is an inner product). Let \(A\) be an \(n \times n\) matrix and \(T\) the linear operator \(T(X) = AX\). Find \(T^*\). If \(Y\) is a fixed element of \(V\), find the element \(Z\) of \(V\) which determines the linear functional \(X \to Y^*X\). In other words, find \(Z\) such that \(Y^*X = (X \mid Z)\) for all \(X\) in \(V\).
Solution
Since \[ (X\mid Y)=Y^*GX \] is an inner product, \(G\) is Hermitian positive definite and in particular invertible.
1. Compute the adjoint of \(T(X)=AX\).
Suppose \(T^*\) is represented by a matrix \(B\). The defining identity is \[ (AX\mid Y)=(X\mid BY) \] for all \(X,Y\). The two sides are \[ Y^*GAX \] and \[ (BY)^*GX=Y^*B^*GX. \] Thus \[ B^*G=GA. \] Multiplying on the right by \(G^{-1}\) gives \[ B^*=GAG^{-1}. \] Taking adjoints and using \(G^*=G\) yields \[ \boxed{B=G^{-1}A^*G.} \] Therefore \[ \boxed{T^*(X)=G^{-1}A^*GX.} \]
2. Represent the functional \(X\mapsto Y^*X\).
We seek \(Z\) such that \[ Y^*X=(X\mid Z)=Z^*GX \] for every \(X\). Hence \[ Z^*G=Y^*. \] Taking adjoints gives \[ GZ=Y, \] so \[ \boxed{Z=G^{-1}Y.} \] Indeed, \[ (X\mid G^{-1}Y) =(G^{-1}Y)^*GX =Y^*X. \]