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Thread: Formally Dual?...?

  1. #1
    Super Member Showcase_22's Avatar
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    Formally Dual?...?

    Consider two finite dimensional vector spaces $\displaystyle U$ and $\displaystyle V$ over complex numbers $\displaystyle \mathbb{C}$, their bases $\displaystyle e_i \in U, \ i=1,2,....,n$ and $\displaystyle f_i \in V, \ i=1,2,....,m$, their dual spaces $\displaystyle U^*$ and $\displaystyle V^*$, and their dual bases $\displaystyle e^i$ and $\displaystyle f^i$.

    Two linear maps $\displaystyle T:U \rightarrow V$ and $\displaystyle S: V \rightarrow U$ are formally dual if:

    $\displaystyle <T(u),v>=<u,S(v)> \ \forall v \in V, \ u \in U$

    Prove that the linear maps $\displaystyle T$ and $\displaystyle T_U^{-1} T^*T_V$ are formally dual.

    $\displaystyle T: U \rightarrow V$
    $\displaystyle T^*: V^* \rightarrow U^*$ so that for each $\displaystyle \alpha \in V^*, \ T^*(\alpha)$ is an element of $\displaystyle U^*$ defined by $\displaystyle T^*(\alpha)(u)=\alpha(T(u)) \ \forall u \in U$.

    $\displaystyle T_U: U \rightarrow U^*$ defined by $\displaystyle T_U(w)(u)=<w,u> \ \forall w,u \in U$ (I know that $\displaystyle T_U(e_i)=e^i$)
    It's probably better to start with the RHS and get the LHS.

    ie. $\displaystyle <T(u),v>=<u,T_U^{-1}T^*T_V(v)>$

    It's also clear that $\displaystyle S=T_U^{-1}T^*T_V:V \rightarrow V^* \rightarrow U^* \rightarrow U$ or $\displaystyle S(v)=T_U^{-1}T^*T_V(v)$

    Take a basis vector of $\displaystyle V$, call it $\displaystyle f_i$. Do the same for $\displaystyle U$, call the basis vector $\displaystyle e_j$.

    I have to get $\displaystyle <T(e_j), f_i>=<e_i,T_U^{-1}T^*T_V(f_i)>$

    We get:

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>$ since $\displaystyle T_U(e_i)=e^i$ and i'm now working on a vector in $\displaystyle V$. The linear map $\displaystyle f^i$ also has to act on a vector, so it's acting on some fixed $\displaystyle v \in V$.

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^* f^i T(u)>$

    Here $\displaystyle f^iT \in U^*$ but linear maps are defined on how they work on a vector so I need to put a $\displaystyle u$ in there.

    Apply $\displaystyle T$ to both sides to get:

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^*f^i T(u)>=<T(e_j), T T_U^{-1}T^*f^i T(u)>$

    From here i'm absolutely lost! Does anyone have any ideas??

    (P.S This is very hard!)
    Last edited by Showcase_22; Nov 25th 2009 at 09:58 AM.
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    It's probably better to start with the RHS and get the LHS.

    ie. $\displaystyle <T(u),v>=<u,T_U^{-1}T^*T_V(v)>$

    It's also clear that $\displaystyle S=T_U^{-1}T^*T_V:V \rightarrow V^* \rightarrow U^* \rightarrow U$ or $\displaystyle S(v)=T_U^{-1}T^*T_V(v)$

    Take a basis vector of $\displaystyle V$, call it $\displaystyle f_i$. Do the same for $\displaystyle U$, call the basis vector $\displaystyle e_j$.

    I have to get $\displaystyle <T(e_j), f_i>=<e_i,T_U^{-1}T^*T_V(f_i)>$

    We get:

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>$ since $\displaystyle T_U(e_i)=e^i$ and i'm now working on a vector in $\displaystyle V$. The linear map $\displaystyle f^i$ also has to act on a vector, so it's acting on some fixed $\displaystyle v \in V$.

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^* f^i T(u)>$

    Here $\displaystyle f^iT \in U^*$ but linear maps are defined on how they work on a vector so I need to put a $\displaystyle u$ in there.

    Apply $\displaystyle T$ to both sides to get:

    $\displaystyle <e_j, S(f_i)>=<e_j, T_U^{-1}T^*f^i(v)>=<e_j, T_U^{-1}T^*f^i T(u)>=<T(e_j), T T_U^{-1}T^*f^i T(u)>$

    From here i'm absolutely lost! Does anyone have any ideas??

    (P.S This is very hard!)

    To begin with, a good idea would be, perhaps, to tell us what are $\displaystyle T_U\,,\,T_V$ ...

    Tonio
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  3. #3
    Senior Member Shanks's Avatar
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    Let U be finite dimention Hilbert space, for any $\displaystyle t \in U$, associates $\displaystyle t^{*} \in U^* $ defined by $\displaystyle
    t^*(x) = < t, x >$. (Define $\displaystyle v^*$ Similarly in V)
    Denote the mapping $\displaystyle {\pi} : U^* --> U$ :
    $\displaystyle v^{*}T=t^*-->t$ ,Since Hilbert space is self-dual,the mapping is injective,
    then RHS
    $\displaystyle =<u,T_{U}^{-1}T^{*}T_{V}(v)>$
    $\displaystyle =<u,T_{U}^{-1}T^{*}(v^{*})>=<u,T_{U}^{-1}(v^{*}T)>$
    $\displaystyle =< T_{U}^{-1}(v^{*}T),u>=<t,u>=t^*(u)$
    $\displaystyle = v^*T(u) = < v, T(u)> = < T(u), v> $
    = LHS
    Last edited by Shanks; Nov 26th 2009 at 08:46 AM. Reason: format
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  4. #4
    Senior Member Shanks's Avatar
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    Adjoint Operator

    For further information, please refer to "Adjoint Operator".
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