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Orthogonal theoretical question

  1. Mar 3, 2009 #1
    V is a space of inner muliplication.
    W1 and W2 are two subspaces of V, so dimW1<dimW2
    prove that there is a vector
    [tex]
    0\neq v\epsilon W_2
    [/tex]
    which is orthogonal to W1
    ??
     
  2. jcsd
  3. Mar 3, 2009 #2
    i got this solution:
    dim W1=k
    dim W2=n

    [tex]n-k=W_2^\perp[/tex]
    i cant see why its correcti dont know that all the vectors of W1 exist in W2 too
    ??
     
    Last edited: Mar 3, 2009
  4. Mar 3, 2009 #3
    Since dim(W1)<dim(W2) you can find a nonzero vector v in W2 which is not an element of W1. Use the orthogonal projection onto W1 to construct, from v, a vector which is orthogonal to W1.
     
  5. Mar 3, 2009 #4
    its a theoretical question i cant construct projections
    the only way is if this "
    dim W1=k
    dim W2=n

    [tex]n-k=W_2^\perp[/tex]
    "

    if all the vectors of W1 exist in W2
    which wasnt gisnt given that
    ??
     
  6. Mar 3, 2009 #5
    What I am trying to tell you is this: If V is an inner product space and W is subspace, then any vector can be written as a sum v=w+n where w is in W and n is in the orthogonal complement of W. In fact, w is the projection of v to W. So, assuming that v is not an element of W, then n will be nonzero and orthogonal to W.
    Usings these facts you will be able to solve the problem.

    You do not need W1 to be a subspace of W2.
     
  7. Mar 4, 2009 #6
    whats
    "V is an inner product space "
    ??
     
    Last edited: Mar 4, 2009
  8. Mar 4, 2009 #7
    i am trying to imagine what you are telling
    but its so abstract
    ??
     
  9. Mar 4, 2009 #8
    suppose we have a four dimentional space
    W={(1,0,0,0),(0,1,0,0)}
    W complement is ={(0,0,1,0),(0,0,0,1)}
    w is subspace of W
    n is subspace of W complement
    v could be written as w+n
    but for that W and W complement must be independant to one another

    we havent been told that
    we have been told that
    dim W2>dim W1

    so there could a case were W2 has all the vectors of W1 and there is another subspace

    so not every vector could be written as a manipulation of W2 and W1
    ??
     
  10. Mar 4, 2009 #9
    A http://en.wikipedia.org/wiki/Inner_product_space" [Broken], I am assuming this is what you mean by "a space of inner multiplication".

    I am not sure what you mean here, I will try to explain it differently.

    What you are given: [tex]W_1[/tex], [tex]W_2[/tex] are subspaces of [tex]V[/tex] with [tex]\dim(W_1)<\dim(W_2)[/tex].
    Let's put [tex]n=\dim(V)[/tex], [tex]n_1=\dim(W_1)[/tex], [tex]n_2=\dim(W_2)[/tex].

    So [tex]\dim(W_1^{\perp}\cap W_2)\ge (n-n_1)+n_2-n\ge 1[/tex].

    But this means that there are nonzero vectors in [tex]W_2[/tex] which are orthogonal to [tex]W_1[/tex].

    I hope you can fill in the details, specifically, why the inequality I wrote is true.
     
    Last edited by a moderator: May 4, 2017
  11. Mar 5, 2009 #10
    i cant understanf how did you constract this equation
    [tex]
    \dim(W_1^{\perp}\cap W_2)\ge (n-n_1)+n_2-n\ge 1
    [/tex]
    if[tex](W_1^{\perp}=n-n_1[/tex] and [tex]n_2=dim(W2)[/tex]
    so the intersect is subtraction of the two groups
    it should be
    [tex]W_1^{\perp}\cap W_2=n-n_1-n_2[/tex]

    and why
    you add =>1
    ??
     
  12. Mar 5, 2009 #11
    I am using the following formula: If A,B are subspaces of V then

    [tex]\dim(A+B)+\dim(A\cap B)=\dim(A)+\dim(B)[/tex]

    and

    [tex]\dim(A+B)\le\dim(V)[/tex]

    If dim(V)>0 then V contains nonzero vectors.


    A general comment: A message board is not the best way to learn linear algebra. You need someone you can talk to face to face, a teacher or other students. If you are stuck on a specific problem, then the people here can give you hints, but this only works if you already know most of what you should know to solve a particular exercise.
     
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