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Math Help - adjoint operator

  1. #1
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    adjoint operator

    I recently teach myself linear algebra with Friedberg's textbook.
    And I have a question about adjoint operator, which is on p.367.

    Definition Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

    Then ,my question is how to prove that there is a unique adjoint T* of T ?

    Can anyone give me some tips ? thanks^^
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  2. #2
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    Quote Originally Posted by typhoonss821 View Post
    I recently teach myself linear algebra with Friedberg's textbook.
    And I have a question about adjoint operator, which is on p.367.

    Definition Let T : V → W be a linear transformation where V and W are finite-dimensional inner product spaces with inner products <‧,‧> and <‧,‧>' respectively. A funtion T* : W → V is called an adjoint of T if <T(x),y>' = <x,T*(x)> for all x in V and y in W.

    Then ,my question is how to prove that there is a unique adjoint T* of T ?

    Can anyone give me some tips ? thanks^^

    Suppose there are two adjoints, T^*\,,\,\,T^{'} to a transf., then for all vectors we get : <Tx,y>=<x,T^*y>=<x,T^{'}y>\Longrightarrow <x,(T^*-T^{'})(y)>= 0  \forall y\,,\,\,(T^*-T^{'})(y)=0\Longrightarrow T^*=T^{'}

    Tonio
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  3. #3
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    That was much simpler than what I was thinking: You know there are 4 main properties of the adjoint; Suppose S, T are linear operators on V and let k \in K then

    1. (S+T)^*=S^*+T^*
    2. (kT)^* = \bar{k}T^*
    3. (ST)^*=T^*S^*
    4. (T*)^*=T

    If you examine all of these properties, you will notice that the uniqueness of the adjoint implies that particiular property. For example \forall u,v \in V , < (S+T)(u), v > = <S(u) +T(u), v >= <S(u), v > + < T(u),v > = < u, S^*(v) > + < u, T^*(v) > = < u, S^*(v) + T^*(v) > = <u, (S^*+T^*)(v) >. You see the uniqueness of the adjoint implies (S+T)*=S*+T*.

    Also, <(kT)(u),v > = < kT(u), v > = k <u, T^*(v) > = < u,\bar{k}T^*(v) > = <u, (\bar{k}T^*)(v)>, again the uniqueness of the adjoint implies (kT)^* = \bar{k}T^*. The same with other two properties, they all show the uniqueness of the adjoint.
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