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Math Help - theorem on adjoint of linear transformation

  1. #1
    Senior Member abhishekkgp's Avatar
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    theorem on adjoint of linear transformation

    Suppose T:V \rightarrow W is a linear transformation, where V and W are finite dimensional inner product spaces. Prove that \text{range } T^{*}=(\text{null } T)^{\perp}.

    I can prove it one direction:
    Let v \in \text{range } T^{*}.

    Then v=T^{*}w for some w \in W.

    Let v_1 \in \text{null } T.

    Then \left<v_1,T^{*}w \right>=\left<Tv_1,w \right>=\left<0,w \right>=0.

    So \left< v_1, v \right>=0 \, \forall \, v_1 \in \text{null } T.

    Hence v \in (\text{null } T)^{\perp}.

    We have \text{range } T^{*} \subseteq (\text{null } T)^{\perp}.


    Can someone show the reverse containment??
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Re: theorem on adjoint of linear transformation

    Quote Originally Posted by abhishekkgp View Post
    Can someone show the reverse containment??
    Hint Use that V,\;W are finite dimensional vector spaces.
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Thumbs up Re: theorem on adjoint of linear transformation

    Quote Originally Posted by FernandoRevilla View Post
    Hint Use that V,\;W are finite dimensional vector spaces.
    Let \{e_1,e_2,\ldots,e_m\} be an orthogonal basis of \text{null }} T.
    Let \{e_1,e_2, \ldots,e_n \} be its extension to an orthogonal basis of V.
    So \{e_{m+1},e_{m+2},\ldots,e_n\} is an orthogonal basis of (\text{null } T)^{\perp}.

    I need to show that \exists \, w \in W \text{ such that } T^{*} w= e_i, \, \text{ where } (m+1)\leq i \leq n.

    Can you please again drop a hint?
    I know i have to use the fact that W is also finite dimensional.
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  4. #4
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    Re: theorem on adjoint of linear transformation

    suppose y \in (\text{image }T^*)^\bot

    so \left<y,T^*w\right> = 0, for all w in W.

    then \left<Ty,w\right> = 0, for all w in W.

    so Ty = 0, so y is in null(T).

    (note: why can we conclude that Ty = 0? well, we can always pick w = Ty, since Ty

    is an element of W, and since <Ty,w> = 0 for all w in W, in particular,

    <Ty,Ty> = 0, so Ty = 0).

    on the other hand, if y is in null(T),

    \left<Ty,w\right> = \left<0,w\right> = 0 for any w in W,

    but \left<Ty,w\right> =\left<y,T^*w\right>

    so we conclude that y \in (\text{image }T^*)^\bot

    so (\text{image }T^*)^\bot = \text{null }T.

    now take the perp of both sides.
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Re: theorem on adjoint of linear transformation

    Quote Originally Posted by Deveno View Post
    suppose y \in (\text{image }T^*)^\bot

    so \left<y,T^*w\right> = 0, for all w in W.

    then \left<Ty,w\right> = 0, for all w in W.

    so Ty = 0, so y is in null(T).

    (note: why can we conclude that Ty = 0? well, we can always pick w = Ty, since Ty

    is an element of W, and since <Ty,w> = 0 for all w in W, in particular,

    <Ty,Ty> = 0, so Ty = 0).

    on the other hand, if y is in null(T),

    \left<Ty,w\right> = \left<0,w\right> = 0 for any w in W,

    but \left<Ty,w\right> =\left<y,T^*w\right>

    so we conclude that y \in (\text{image }T^*)^\bot

    so (\text{image }T^*)^\bot = \text{null }T.

    now take the perp of both sides.
    thank you Denevo. that helped.
    But i am still looking for a direct way to solve the problem.
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