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Thread: orthogonal subset proof

  1. #1
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    orthogonal subset proof

    A,B are subsets of$\displaystyle R^{n}$prove that if$\displaystyle A^{\perp}\subset B^{\perp}$then there is $\displaystyle v\epsilon A$ so v is not part of SpB
    how i tryed to solve it:
    $\displaystyle s\epsilon A^{\perp}$
    $\displaystyle t\epsilon B^{\perp}$
    if there is $\displaystyle v\epsilon A$ then $\displaystyle v\bullet s=0$ and because $\displaystyle A^{\perp}\subset B^{\perp}$ then $\displaystyle v\bullet t=0$
    so because v is othogonal to $\displaystyle B^{\perp}$and to $\displaystyle A^{\perp}$so $\displaystyle
    SpA\subseteq SpB
    A^{\perp}=(SpA){}^{\perp}
    $
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  2. #2
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    re: orthogonal subset proof

    No. You're not correctly grasping the implications of the containment $\displaystyle A^{\perp}\subsetneq B^{\perp}$.

    We know that there exists a vector $\displaystyle u\in B^{\perp}\setminus A^{\perp}$. This means that $\displaystyle u$ is orthogonal to every element of $\displaystyle B$, but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to.

    Can you prove it from here?
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  3. #3
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    re: orthogonal subset proof

    Quote Originally Posted by topspin1617 View Post
    No. You're not correctly grasping the implications of the containment $\displaystyle A^{\perp}\subsetneq B^{\perp}$.

    We know that there exists a vector $\displaystyle u\in B^{\perp}\setminus A^{\perp}$. This means that $\displaystyle u$ is orthogonal to every element of $\displaystyle B$, but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to.

    Can you prove it from here?
    i cant undrstand this term
    $\displaystyle A^{\perp}\subsetneq B^{\perp}$
    if $\displaystyle B^{\perp}$={(1,0,0),(0,1,0)}
    and $\displaystyle A^{\perp}$={(1,0,0)}
    then
    $\displaystyle A^{\perp}\subsetneq B^{\perp}$
    so (0,0,1) orthogonal to $\displaystyle A^{\perp}$ and to $\displaystyle B^{\perp}$
    so i cant see how

    "but that there exists an element of $\displaystyle A$ which $\displaystyle u$ is NOT orthogonal to."

    is possible
    ?
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  4. #4
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    re: orthogonal subset proof

    Did you read my entire post? Since $\displaystyle A^{\perp}$ is properly contained in $\displaystyle B^{\perp}$, there must exist an element $\displaystyle u$ which is in $\displaystyle B^{\perp}$ but not in $\displaystyle A^{\perp}$. Correct?

    Now, consider this vector $\displaystyle u$. What does it mean to say $\displaystyle u\in B^{\perp}$? What does it mean to say $\displaystyle u\notin A^{\perp}$?
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  5. #5
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    re: orthogonal subset proof

    Do you see the little "/" in $\displaystyle \subsetneq$ (I didn't at first)? That removes the possibility that A= B.

    $\displaystyle A\subseteq B$ means that A is a subset that might be equal to B.

    $\displaystyle A\subsetneq B$ means that A is a proper subset of B- it is NOT equal to B.

    $\displaystyle A\subset B$ is ambiguous but generally should be considered the same as $\displaystyle A \subseteq B$.
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  6. #6
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    re: orthogonal subset proof

    aahhhh correct
    i understood this point
    now i think about the profe
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  7. #7
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    re: orthogonal subset proof

    Quote Originally Posted by topspin1617 View Post
    Did you read my entire post? Since $\displaystyle A^{\perp}$ is properly contained in $\displaystyle B^{\perp}$, there must exist an element $\displaystyle u$ which is in $\displaystyle B^{\perp}$ but not in $\displaystyle A^{\perp}$. Correct?

    Now, consider this vector $\displaystyle u$. What does it mean to say $\displaystyle u\in B^{\perp}$? What does it mean to say $\displaystyle u\notin A^{\perp}$?
    i understand that there are members in $\displaystyle B^{\perp}$ which are not in $\displaystyle A^{\perp}$
    so not every vector in $\displaystyle B^{\perp}$ orthogonal to A

    but i dont know how to prove that there is a vector who is in A but not in Sp B


    i tried some thing
    dim($\displaystyle B^{\perp}$)+dim SpB= n
    dim($\displaystyle A^{\perp}$)+dim SpA= n
    so
    dim($\displaystyle A^{\perp}$)+dim SpA=dim$\displaystyle B^{\perp}$+dim SpB
    so if $\displaystyle A^{\perp}\subset B^{\perp}$ then dim$\displaystyle A^{\perp}$<dimdim$\displaystyle B^{\perp}$

    so dim SpA>dim SpB
    what now?
    Last edited by transgalactic; Jun 25th 2011 at 07:57 AM.
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