Let \(V\) be a finite-dimensional inner product space and \(T\) a positive linear operator on \(V\). Let \(p_T\) be the inner product on \(V\) defined by \(p_T(\alpha, \beta) = (T\alpha \mid \beta)\). Let \(U\) be a linear operator on \(V\) and \(U^*\) its adjoint with respect to \(( \mid )\). Prove that \(U\) is unitary with respect to the inner product \(p_T\) if and only if \(T = U^*TU\).
Solution
1. Express preservation of \(p_T\) in terms of the original inner product.
By definition, \(U\) is unitary for \(p_T\) exactly when \[ p_T(U\alpha,U\beta)=p_T(\alpha,\beta) \] for all \(\alpha,\beta\in V\). Using \(p_T(\alpha,\beta)=(T\alpha\mid\beta)\), this is \[ (TU\alpha\mid U\beta)=(T\alpha\mid\beta) \] for all \(\alpha,\beta\).
2. Move \(U\) from the second slot by means of the adjoint.
For arbitrary \(x,y\in V\), \[ (x\mid Uy)=(U^*x\mid y). \] Apply this with \(x=TU\alpha\) and \(y=\beta\). The condition from step 1 becomes \[ (U^*TU\alpha\mid\beta)=(T\alpha\mid\beta) \] for every \(\alpha,\beta\).
3. Equality of all matrix coefficients is equivalent to equality of the operators.
If \[ (U^*TU\alpha-T\alpha\mid\beta)=0 \] for every \(\beta\), then taking \(\beta=U^*TU\alpha-T\alpha\) gives \[ \|U^*TU\alpha-T\alpha\|^2=0. \] Hence \[ U^*TU\alpha=T\alpha \] for every \(\alpha\), so \[ U^*TU=T. \] The argument reverses verbatim: if \(U^*TU=T\), then step 2 gives \(p_T(U\alpha,U\beta)=p_T(\alpha,\beta)\) for all \(\alpha,\beta\).
Therefore \[ \boxed{U\text{ is unitary for }p_T\iff T=U^*TU.} \]