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Prove a set is not a vector space

  1. Aug 8, 2014 #1
    1. The problem statement, all variables and given/known data

    Let [itex]b[/itex] be a symetric bilinear form on [itex]V[/itex] and [itex]A = \{ v\in V : b\left(v,v\right)=0\}[/itex]. Prove that [itex]A[/itex] is not a vector space, unless [itex]A = 0[/itex] or [itex]A = V[/itex].

    2. The attempt at a solution

    If we suppose that [itex]A[/itex] is a vector space then for every [itex]v,w\in A[/itex] we must have: [itex]b\left(v+w,v+w\right) = b\left(v,w\right)[/itex]. This try didn't go anywhere. I think I should construct a counter example, but I wouldn't know from where to start.
     
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  3. Aug 8, 2014 #2

    Orodruin

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    Assume that ##V = \mathbb R^2## and that
    $$
    b(v,w) =
    v^T
    \begin{pmatrix}
    1 & 0 \\ 0 & 0
    \end{pmatrix}
    w.
    $$
    For ##v = \begin{pmatrix} x \\ y\end{pmatrix}##, we now have
    $$
    b(v,v) = x^2
    $$
    so ##A## is the set of vectors with ##x = 0##, i.e., vectors of the form
    $$
    v = \begin{pmatrix} 0 \\ y \end{pmatrix},
    $$
    which is basically the vector space ##\mathbb R##. Thus ##A \neq 0## and ##A \neq V## while still being a vector space.

    Are you sure the problem formulation is correct?
     
  4. Aug 8, 2014 #3
    I think so. It's problem 12, page 53 from O' Niell's book "Semi-Riemannian Geometry with Applications to General Relativity".
     
  5. Aug 8, 2014 #4

    Orodruin

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    I do not have that book but "Semi-Riemannian" suggests that eventually there will be a (pseudo-Riemannian) metric involved. There were no additional constraints on the bilinear form apart from just being symmetric? For example, the light-cone extending from the origin of Minkowski space is obviously not a vector space as addition of vectors on the light-cone can end up being time-like or space-like.
     
  6. Aug 8, 2014 #5
    No. V is an finitely dimensional real vector space over the real numbers and b is just a bilinear form. It doesn't mention semi-Riemannian or even Minkowski spaces.

    Maybe the author made a mistake...
     
  7. Aug 9, 2014 #6

    vela

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    I ran across a thread on stackexchange which suggests that the problem should have said that ##A## is not a vector space unless ##A=0## or ##A=N##, where ##N## is the nullspace of ##b##.
     
  8. Aug 10, 2014 #7

    pasmith

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    In that case, you can write [itex]b(v,w) = v^TBw[/itex] for some real symmetric matrix [itex]B[/itex].

    It is then easy to show that [itex]\ker B \subset A[/itex]. But [itex]A \subset \ker B[/itex] doesn't necessarily hold.

    Consider the case where [itex]V = \mathbb{R}^4[/itex] and [itex]B[/itex] is diagonal with eigenvalues 1, 1, -1 and 0 so that [tex]
    v^TBw = v_1w_1 + v_2w_2 - v_3w_3.
    [/tex] Now [itex]\ker B = \{(0,0,0,t) : t \in \mathbb{R} \}[/itex] and [itex]A = \{(x,y,z,t) \in \mathbb{R}^4 : z^2 = x^2 + y^2\}[/itex] which is the product of a double cone and the real line. It's not a subspace of [itex]\mathbb{R}^4[/itex].

    If it happens that [itex]A = \ker B[/itex] then [itex]A[/itex] is indeed a subspace. That [itex]A = \ker B[/itex] need not be [itex]\{0\}[/itex] or [itex]V[/itex] is shown by Orodruin's example.
     
  9. Aug 14, 2014 #8

    pasmith

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    And since [itex]B[/itex] is symmetric it is diagonalisable, so there exists a basis of eigenvectors. There is no reason not to use that basis, so you can take [tex]
    b(v,w) = \sum_{i=1}^n \lambda_i v_i w_i
    [/tex] where [itex]v = \sum v_i e_i[/itex], [itex]w = \sum w_i e_i[/itex] and [itex]e_i[/itex] is an eigenvector of [itex]B[/itex] with eigenvalue [itex]\lambda_i[/itex].

    Your aim is to show that if [itex]v \in A \setminus \ker B[/itex] then there exists a [itex]w \in A \setminus \ker B[/itex] such that at least one of [itex]v + w[/itex] and [itex]v - w[/itex] is not in [itex]A[/itex].
     
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