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,
- (S + T)* = S* + T*
- (aT)* = `aT*
- (ST)* = T*S*
- (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