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Math Help - Formally Dual?...?

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

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

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

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

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

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

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

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

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

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

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

    We get:

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

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

    Apply T to both sides to get:

    <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; November 25th 2009 at 10:58 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. <T(u),v>=<u,T_U^{-1}T^*T_V(v)>

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

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

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

    We get:

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

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

    Apply T to both sides to get:

    <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 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 t \in U, associates t^{*} \in U^* defined by <br />
t^*(x) = < t, x >. (Define v^* Similarly in V)
    Denote the mapping {\pi} : U^* --> U :
     v^{*}T=t^*-->t ,Since Hilbert space is self-dual,the mapping is injective,
    then RHS
    =<u,T_{U}^{-1}T^{*}T_{V}(v)>
    =<u,T_{U}^{-1}T^{*}(v^{*})>=<u,T_{U}^{-1}(v^{*}T)>
    =< T_{U}^{-1}(v^{*}T),u>=<t,u>=t^*(u)
    = v^*T(u) = < v, T(u)> = < T(u), v>
    = LHS
    Last edited by Shanks; November 26th 2009 at 09:46 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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