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Prove that the set of all vectors is a subspace

  1. Jul 9, 2008 #1

    danago

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    Let [tex]\vec{a} \ne 0[/tex] be a fixed vector in R3.

    (a) Prove that the set of all vectors [tex]\vec{x} \in R^3[/tex] satisfying [tex]\vec{a}.\vec{x}=0[/tex] is a subspace of R^3. Describe this set geometrically.

    (b) Is the set of all vectors [tex]\vec{x} \in R^3[/tex] satisfying [tex]\vec{a} \times \vec{x}=0[/tex] a subspace of R3?



    For part (a), i thought about the problem geometrically and id be inclined to say that it is a subspace. The zero dot product implies that a and x are orthogonal, and so the set of all x will be a plane through the origin (since <0,0>.a=0 for any a) which is perpendicular to a.

    Analytically, i proved it by letting p and q be vectors such that p.a=q.a=0. I then showed that any linear combination of these two vectors is also perpendicular to a.

    (kp+lq).a=k(p.a)+l(q.a)=0

    Is that correct?

    Its part (b) where im a bit stuck. If the cross product of two vectors is the zero vector, this implies that they are parallel, yea? So x is some element of the set of all vectors which are parallel to a? Since the linear combination of any two vectors parallel to a will also be parallel to a, can we conclude that the set of all x is a subspace of R3?

    Now, if my reasoning is correct, how could i go about showing this analytically? I guess it comes down to showing that the cross product being zero implies that the vectors are parallel; once i can show this i can do the rest myself.

    Thanks in advance,
    Dan.
     
  2. jcsd
  3. Jul 9, 2008 #2

    Defennder

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    Re: Subspaces

    You can do (b) the same way you did (a). Just use the properties of cross product to do so.
     
  4. Jul 9, 2008 #3

    danago

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    Re: Subspaces

    Oh...ofcourse..haha dont know why i didnt do that. Guess its time for bed :tongue2: Thanks for the quick reply :smile:
     
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