Results 1 to 2 of 2

Thread: Bilinear maps and Dual Vector Spaces

  1. #1
    Super Member Showcase_22's Avatar
    Joined
    Sep 2006
    From
    The raggedy edge.
    Posts
    782

    Bilinear maps and Dual Vector Spaces

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

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

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

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

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

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

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

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

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

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

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

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

    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
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    Quote Originally Posted by Showcase_22 View Post
    Let $\displaystyle \tau: W \times V \rightarrow \mathbb{F}$ be a bilinear map, where $\displaystyle V$ and $\displaystyle W$ are vector spaces over a field $\displaystyle \mathbb{F}$. Recall that the dual vector space is denoted by $\displaystyle V^*$. Let $\displaystyle U, \ U_i$ be subspaces of V.

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

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

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

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

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

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

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


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


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

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

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

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

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


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

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

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

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

    Tonio

    I'd really appreciate any help with this. I'm finding the concept of bilinear maps and dual vector spaces quite hard =S
    .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. vector spaces and linear maps
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Nov 26th 2011, 08:30 AM
  2. [SOLVED] Bilinear map and Dual space
    Posted in the Advanced Algebra Forum
    Replies: 11
    Last Post: Nov 16th 2010, 10:40 AM
  3. Bilinear/Linear Maps
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Mar 30th 2010, 02:30 PM
  4. Bilinear Maps
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: Sep 21st 2009, 06:32 PM
  5. Dual Spaces
    Posted in the Advanced Algebra Forum
    Replies: 0
    Last Post: Apr 21st 2009, 09:38 AM

Search Tags


/mathhelpforum @mathhelpforum