Results 1 to 7 of 7

Math Help - orthogonal subset proof

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    orthogonal subset proof

    A,B are subsets of  R^{n}prove that if  A^{\perp}\subset B^{\perp}then there is v\epsilon A so v is not part of SpB
    how i tryed to solve it:
     s\epsilon A^{\perp}
    t\epsilon B^{\perp}
    if there is v\epsilon A then v\bullet s=0 and because A^{\perp}\subset B^{\perp} then v\bullet t=0
    so because v is othogonal to B^{\perp}and to A^{\perp}so  <br />
SpA\subseteq SpB<br />
A^{\perp}=(SpA){}^{\perp}<br />
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Nov 2010
    Posts
    193

    re: orthogonal subset proof

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

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

    Can you prove it from here?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: orthogonal subset proof

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

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

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

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

    is possible
    ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Nov 2010
    Posts
    193

    re: orthogonal subset proof

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

    Now, consider this vector u. What does it mean to say u\in B^{\perp}? What does it mean to say u\notin A^{\perp}?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,423
    Thanks
    1331

    re: orthogonal subset proof

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

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

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

    A\subset B is ambiguous but generally should be considered the same as A \subseteq B.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: orthogonal subset proof

    aahhhh correct
    i understood this point
    now i think about the profe
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    re: orthogonal subset proof

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

    Now, consider this vector u. What does it mean to say u\in B^{\perp}? What does it mean to say u\notin A^{\perp}?
    i understand that there are members in B^{\perp} which are not in A^{\perp}
    so not every vector in 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( B^{\perp})+dim SpB= n
    dim( A^{\perp})+dim SpA= n
    so
    dim( A^{\perp})+dim SpA=dim B^{\perp}+dim SpB
    so if  A^{\perp}\subset B^{\perp} then dim A^{\perp}<dimdim B^{\perp}

    so dim SpA>dim SpB
    what now?
    Last edited by transgalactic; June 25th 2011 at 07:57 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Proof that if A is a subset of P(A), P(A) is a subset of P(P(A))
    Posted in the Discrete Math Forum
    Replies: 12
    Last Post: May 23rd 2011, 04:26 PM
  2. Proof on Openness of a Subset and a Function of This Subset
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: October 24th 2010, 09:04 PM
  3. proof of closed subset
    Posted in the Differential Geometry Forum
    Replies: 13
    Last Post: September 5th 2010, 03:51 AM
  4. proof about a subset
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: October 12th 2009, 09:21 AM
  5. Aother subset proof
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: February 6th 2007, 02:40 PM

Search Tags


/mathhelpforum @mathhelpforum