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Thread: theorem on adjoint of linear transformation

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

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

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

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

    Let $\displaystyle v_1 \in \text{null } T$.

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

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

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

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


    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 $\displaystyle V,\;W$ 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 $\displaystyle V,\;W$ are finite dimensional vector spaces.
    Let $\displaystyle \{e_1,e_2,\ldots,e_m\}$ be an orthogonal basis of $\displaystyle \text{null }} T$.
    Let $\displaystyle \{e_1,e_2, \ldots,e_n \}$ be its extension to an orthogonal basis of $\displaystyle V$.
    So $\displaystyle \{e_{m+1},e_{m+2},\ldots,e_n\}$ is an orthogonal basis of $\displaystyle (\text{null } T)^{\perp}$.

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

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

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

    then $\displaystyle \left<Ty,w\right> = 0$, 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),

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

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

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

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

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

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

    then $\displaystyle \left<Ty,w\right> = 0$, 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),

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

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

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

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

    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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