 Solution

Use the convention that the inner product is conjugate-linear in its first argument and linear in its second, as in part (b). The adjoint operator is characterized byExistence and uniqueness follow from the finite-dimensional representation of linear functionals by the inner product. More concretely, choose an orthonormal basis $e_i$ and let $a_{ij}=⟨e_i,T{e}_j⟩$. Thenwhich defines $T^{\ast }$ uniquely and verifies the required identity by expansion in that basis. Thus its matrix is the conjugate transpose, proved here rather than assumed. A self-adjoint operator satisfies $T^{\ast }=T$, equivalently $⟨Tv,w⟩=⟨v,Tw⟩$ for every pair of vectors.