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Linear subspace proof

  1. Aug 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Let V and W be vector spaces over [itex]F [/itex] and [itex]T:V \rightarrow W[/itex] a linear transformation. Prove that [itex]ker(T):=[/itex]{[itex]\epsilon V\mid T()=0_{v}[/itex]} is a vector subspace of [itex]V[/itex]

    2. Relevant equations



    3. The attempt at a solution

    Is it allright just to state the trivial solution.

    ie There exists the vector [itex]0v\epsilon V[/itex] such that
    [itex]T(0v) \rightarrow W_{0}[/itex]

    therfore the vector [itex]0v\epsilon V [/itex] is also [itex]0v\epsilon T [/itex]

    or do I need more Axioms like

    There exists the vectors [itex]-v\epsilon V[/itex] and [itex]v\epsilon V[/itex] such that
    [itex]T(-v+v) = T(0v) \rightarrow W_{0}[/itex]


    to prove that T() is a vector subspace of V
    regards
    Brendan
     
  2. jcsd
  3. Aug 17, 2009 #2
    You're trying to prove that ker(T) is a subspace, so you have to show that 0 is in ker(T). Moreover, you have to show that if u and v are in ker(T), then so is u + v. Finally, if u is in ker(T) and c is a scalar, then you have to show that c*u is in ker(T). These all follow directly from the basic properties of linear transformations.
     
  4. Aug 17, 2009 #3
    Thanks for your help.

    As a subspace always has the zero vector can I just say that for
    {u,v} both elemants of V.
    We have
    0v = A
    0u = B

    Ax= 0 and Ay = 0, then A(x + y) = vx + vy = 0 + 0 = 0

    Ax = 0 and c is a scalar, then A(cx) = cAx = c0 = 0

    {Ax ,Bu} = 0 and c is a scalar, then Acx+Bcy = cAx+cBy = c0 + c0 = 0+0 = 0

    Is that allright?



    Ax = 0 and c is a scalar, then A(cx) = cAx = c0 = 0
     
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