Lecture Vi: Self-adjoint and Unitary Operators

De nition. Let (V, 〈 , 〉) be a n-dimensional euclidean vector space and T : V −→ V a linear operator. We will call the adjoint of T , the linear operator S : V −→ V such that: 〈T (u), v〉 = 〈u, S(v)〉 , for all u, v ∈ V . Proposition 1. Let (V, 〈 , 〉) be a n-dimensional euclidean vector space and T : V −→ V a linear operator. The adjoint of T exists and is unique. Moreover, if E denotes an orthonormal basis of V (with respect to 〈 , 〉) and T has matrix B with respect to E (i.e., T (E) = EB), then the adjoint of T is the linear operator S : V −→ V , that has matrix B with respect to E. For this reason, the adjoint of T is sometimes called the transpose operator of T and denoted T . Proof. Let E an arbitrary basis of V , let T (E) = EB, with B ∈Mn(R) and let A be the matrix of 〈 , 〉 (w.r.t. E). Therefore: 〈u, v〉 = 〈Ex,Ey〉 = xAy for all u = Ex, v = Ey ∈ V [observe that A ∈ GLn(R), since 〈 , 〉 is an inner product]. Let us denote with S : V −→ V an arbitrary linear operator with matrix C ∈ Mn(R) (w.r.t. E), i.e., S(E) = EC. For any u, v ∈ V we have: 〈T (u), v〉 = 〈T (Ex),Ey〉 = 〈EBx,Ey〉 = (Bx)Ay = x (BA)y 〈u, S(v)〉 = 〈Ex, S(Ey)〉 = 〈Ex,ECy〉 = xA(Cy) = x (AC)y .