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Vector space proof

  1. Mar 30, 2004 #1
    how do you prove that if v is an element of V (a vector space), and if r is a scalar and if rv = 0, then either r = 0 or v = 0... it seems obvious, but i have no idea how to prove it...
     
  2. jcsd
  3. Mar 30, 2004 #2
    If v is the vector (a, b, c), then the vector that results from the multiplication rv is (ra, rb, rc). If the result is equal to 0, the zero vector, then (ra, rb, rc) = (0, 0, 0). If we write this formally we get:
    [tex]ra = 0[/tex]
    [tex]rb = 0[/tex]
    [tex]rc = 0[/tex]
    The solution is that either r equals 0, or a, b, and c all equal 0. And if a, b, and c are 0 then the vector v is (0, 0, 0) which is the zero vector.

    Is this proof satisfactory? There are probably a lot of ways to prove this. :smile:
     
  4. Mar 30, 2004 #3

    matt grime

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    That proof requires you to pick a basis. If I pick a different basis, do you know that it still holds?

    Here's the basis free proof. Suppose rv=0, then if r is zero we are done, if not multiply rv=0 by 1/r and we see v=0.
     
  5. Mar 30, 2004 #4
    Yes... if you pick your basis at (A, B, C) then the vector v becomes (a - A, b - B, c - C) and the zero vector is now (A, B, C).
    [tex]r(a - A) = A[/tex]
    [tex]r(b - B) = B[/tex]
    [tex]r(c - C) = C[/tex]
    For this to be true for r <> 0, a must equal A, b must equal B and c must qual C, and thus the vector v becomes (A, B, C) which is again the zero vector.
     
  6. Mar 30, 2004 #5

    matt grime

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    No, that isn't how one does a change of basis (of a vector space: the origin isn't fixed.)
     
  7. Mar 30, 2004 #6

    matt grime

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    It's also dimension dependent. The result is true for every vector space, even those where picking a basis, never mind solving an uncountable set of linear equations requires the axiom of choice.
     
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