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On proving real vector spaces (subspaces)

  1. Jan 31, 2004 #1
    I hope someone can help me (guide) in this theorem.

    How can I show that a "subset W of a vector space V is indeed
    a subspace of V if and only if given u and v as vectors in W and
    a and b are said to be scalars, then au + bv is in W."?

    Can I assume a vector with my desired number of elements?
    Also I am sure that W is a subset of V because it's given.
     
  2. jcsd
  3. Jan 31, 2004 #2

    HallsofIvy

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    What do you mean by "number of elements"? The vector space is not stated to have a particular dimension (and in fact, the statement is true for infinite dimensional vectors spaces as well) so I wouldn't recommend trying to do this by looking at components.

    What is the definition of "subspace". What you need to do is show that if "given u and v as vectors in W and
    a and b are said to be scalars, then au + bv is in W", then all of the properties of the definition of subspace are satisfied.
    (Hint: many of them, such as commutativity of addition, follow from the fact that u, v are in the V and those are true for V.)

    You will also need to show(since this is "if and only if" that, IF W is a subspace of V, THEN "given u and v as vectors in W and a and b are said to be scalars, then au + bv is in W" but that's much easier.
     
  4. Jan 31, 2004 #3
    Hallsof Ivy

    Hello there.

    Yeah, what I mean about the "number of elements" is that the number of components in a vector. But anyway, ... it may not be the concern.

    So au + bv is in W... does it mean that au + bv = bv + au or
    au + (bv + cw) = (au + bv) + cw for w as a vector and c as a constant?

    How about the converse of that statement? WHat is the approach?
     
  5. Feb 2, 2004 #4

    HallsofIvy

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    Yes, of course. au+bv is in W so it is in V. au+ bv= bv+ au is true because V is a vector space and so addition is commutative.
    au, bv, and cw are in W so they are in V. au + (bv + cw) = (au + bv) + cw because addition is associative in V.

    (By the way, a, b, and c are "scalars" or "numbers". They are not necessarily "constant".)
     
  6. Feb 2, 2004 #5
    Thank you very much

    Hello HallsofIvy!

    Now, I get it... =) thank you. PArdon me for the thread on

    the resistances. I didn't take a glance at it and I just

    simply placed it there without knowing that it was easy.
     
  7. Mar 12, 2009 #6
    i want to similar question...
    the question is shown as below...
    Explain why the set w={(x,y)€R^2;|x|=|y|}, is not a real subspace.
    anyone can help me??
    thanks a lot...
     
  8. Mar 12, 2009 #7
    Consider vectors
    [tex] (2,2), (-1,1) \in W [/tex]
    But
    [tex] (2,2)+ (-1,1) = (1,3) [/tex]
    is not in W. Hence not a vector space.
     
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