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Math Help - Bilinear maps and Dual Vector Spaces

  1. #1
    Super Member Showcase_22's Avatar
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    Bilinear maps and Dual Vector Spaces

    Let \tau: W \times V \rightarrow \mathbb{F} be a bilinear map, where V and W are vector spaces over a field \mathbb{F}. Recall that the dual vector space is denoted by V^*. Let U, \ U_i be subspaces of V.

    i). Prove that U^{\perp}=\{w \in W | \tau(w,u)=0 \ \forall u \in U \}

    My answer: w= \alpha_1 w_1+\alpha_2 w_2).
    Hence \tau(\alpha_1 w_1+\alpha_2 w_2,u)=0 by the definition of U^{\perp}.

    We also have that \alpha_1 \tau (w_1,u)+\alpha_2 \tau(w_2,u)=0 from the definition of U^{\perp}.

    Hence we have that \tau(\alpha_1 w_1+ \alpha_2 w_2,u)=\alpha_1 \tau (w_1,u)+ \alpha_2 \tau (w_2,u) so U^{perp} is a subspace of W.

    ii). Prove that U \subset (U^{\perp})^{\perp}.

    This is where things start to go a bit wrong. I have no idea what (U^{\perp})^{\perp} is defined to be!

    iii). Prove that U_1 \subset U_2 \Rightarrow U_2^{\perp} \subset U_1^{\perp}

    I'm also a little stuck on this as well. I decided to start by letting u_1 \in U_1 \Rightarrow u_1 \in U_2.

    I also know that \tau(w,u_1)=0.

    However, I wasn't sure where this would take me =S

    iv). Given that T_U: W \rightarrow U^* defined by T_U(w)(u)=\tau(w,u) for w \in W, \ u \in U is a linear map from W to U^*, and that ker(T_U)=U^{\perp}, deduce that dim(U)+dim(U^{\perp}) \geq dim(W)

    I figure this has something to do with rank+ nullity....

    I'd really appreciate any help with this. I'm finding the concept of bilinear maps and dual vector spaces quite hard =S
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  2. #2
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    Quote Originally Posted by Showcase_22 View Post
    Let \tau: W \times V \rightarrow \mathbb{F} be a bilinear map, where V and W are vector spaces over a field \mathbb{F}. Recall that the dual vector space is denoted by V^*. Let U, \ U_i be subspaces of V.

    i). Prove that U^{\perp}=\{w \in W | \tau(w,u)=0 \ \forall u \in U \}

    Prove what? You only wrote the definition of orthogonal complement. Perhaps it was to prove that this complement is a subspace?

    My answer: w= \alpha_1 w_1+\alpha_2 w_2).
    Hence \tau(\alpha_1 w_1+\alpha_2 w_2,u)=0 by the definition of U^{\perp}.

    We also have that \alpha_1 \tau (w_1,u)+\alpha_2 \tau(w_2,u)=0 from the definition of U^{\perp}.

    Hence we have that \tau(\alpha_1 w_1+ \alpha_2 w_2,u)=\alpha_1 \tau (w_1,u)+ \alpha_2 \tau (w_2,u) so U^{perp} is a subspace of W.

    ii). Prove that U \subset (U^{\perp})^{\perp}.

    This is where things start to go a bit wrong. I have no idea what (U^{\perp})^{\perp} is defined to be!


    == \left(U^\perp \right)^\perp:=\left\{v\in V\;\slash \;\tau(v,w)=0\;\;\forall w\in U^\perp\right\}==


    iii). Prove that U_1 \subset U_2 \Rightarrow U_2^{\perp} \subset U_1^{\perp}

    I'm also a little stuck on this as well. I decided to start by letting u_1 \in U_1 \Rightarrow u_1 \in U_2.

    I also know that \tau(w,u_1)=0.

    However, I wasn't sure where this would take me =S

    \mbox{ Let} w_2\in U_2^\perp \Longrightarrow \forall u_2\in U_2\,,\;\tau(u_2,w_2)=0 , and this is true is particular for all u_1\in U_1 since U_1 \subset U_2 !


    iv). Given that T_U: W \rightarrow U^* defined by T_U(w)(u)=\tau(w,u) for w \in W, \ u \in U is a linear map from W to U^*, and that ker(T_U)=U^{\perp}, deduce that dim(U)+dim(U^{\perp}) \geq dim(W)

    I figure this has something to do with rank+ nullity....

    Of course! And it also has to do with the important fact that U^*\cong U for finite dimensional spaces, so by the rank+nullity theorem:

    \dim W = \dim(\ker T_U=U^\perp)+\dim (Im\,\, T_U) \geq \dim U^\perp +\dim U^*

    Tonio

    I'd really appreciate any help with this. I'm finding the concept of bilinear maps and dual vector spaces quite hard =S
    .
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