Let H1, H2 be Hilbert spaces and

let T : H1 –> H2 be a bounded linear operator.

Def.: The adjoint operator of T is the operator

T*: H1 –> H2 such that

Tx, y > = < x, T*y > , all x of H1, y of H2

This is proved by computing the right side, using

Tx, y > = < x, Ty >

= < Ty, x >,

where the first equality holds because T is self-adjoint and the second equality holds because we are working on a real inner-product space.

If  <Tx, x > =  0 for all x of H1, then we can imply that < Tx, y > =  0 for all x of H1, y of H2 .

This implies that T = 0 (take y = Tx).

An operator on an inner-product space is called normal if it commutes with its adjoint. In the other words, T of L(H) is normal if  TT* = T*T.

Proof: Let T of L(H) We will prove both directions of this result

at the same time. Note that

T is normal     <=> T*T TT* = 0

                     <=> (T*T-TT*)x,x > = 0 , for all x of H

<=> < T*Tx,x > = < TT*x,x > , for all x of H

<=> ll Tx ll² = ll T*x ll² , for all x of H

 where we used < Tx, x > =  0 to establish the second equivalence (note that the operator T*T TT*  is self-adjoint).

Ex.1: If a basis in Cn is given and a linear operator on Cn is represented by a matrix, then its adjoint is represented by the conjugate transpose of that matrix.

Proof: Let V = C² (over C) and linear operators T on V, let T : V -> V be defined by the mapping

T(z1, z2) = (2z1 + iz2, (1-i)z1).

                Evaluate T* at x = (3-i, 1+2i).

In the other words:

Th.: Let H1, H2 be Hilbert spaces and S, T bounded linear operators from H1 to H2. Then,

  1. (S + T)* = S* + T*
  2. (aT)* = `aT*
  3. (ST)* = T*S*
  4. (T*)* = T.

Ex.2: Let T be a linear operator on an inner product space V. let U1 = T + T* and U2 = TT*. Prove that U1 = U*1 and U2 = U*2.

Solution: In two lines,

U*1 = (T+T*)* = T* + (T*)* = T* + T = U1,

U*2 = (TT*)* = (T*)*(T)* = TT* = U2