Let *H*_{1}, *H*_{2} be Hilbert spaces and

let *T* : *H*_{1} –> *H*_{2} be a bounded linear operator.

Def.: The *adjoint operator* of *T* is the operator

*T*^{*}: *H*_{1} –> *H*_{2} such that

< *Tx*, *y >* = < *x*, *T ^{*}y > *, all

*x*of

*H*

_{1},

*y*of

*H*

_{2}

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 *H*_{1}, then we can imply that < *Tx, y >** = ** *0 for all *x* of *H*_{1}, *y* of *H*_{2} .

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 C* ^{n}* is given and a linear operator on C

*is represented by a matrix, then its adjoint is represented by the conjugate transpose of that matrix.*

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

T(z_{1}, z_{2}) = (2z_{1} + iz_{2}, (1-i)z_{1}).

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

In the other words:

Th.: Let *H*_{1}, *H*_{2} be Hilbert spaces and *S, T* bounded linear operators from *H*_{1} to *H*_{2}. 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 U_{1} = T + T* and U_{2} = TT*. Prove that U_{1} = U*_{1} and U_{2} = U*_{2}.

Solution: In two lines,

U*_{1} = (T+T*)* = T* + (T*)* = T* + T = U_{1},

U*_{2} = (TT*)* = (T*)*(T)* = TT* = U_{2}

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