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Possible mistake in question?

  1. Dec 21, 2008 #1
    probably not but if not i dont know where to start...
    let U and W be vector subspaces of V. Show that UnW is a vector subspace of V
    define the subspace U+W. What does it mean to say that V is a direct sum, UdirectsumW of U and W

    understand the direct sum part but surely i need values of U and W to show UnW is a vector space of V?
  2. jcsd
  3. Dec 21, 2008 #2


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    The same way you show that any subset of a vector space is a subspace: show that it is closed under addition of vectors and scalar multiplication.

    If u and v are vectors in UnW then, by definition of "intersection" they are in both U and W. Since u and v are both in U, and U is a subspace of V, u+ v is in U ...

    Can you finish that?
  4. Dec 21, 2008 #3
    ye chears that helps alot, could you take a look at my other post please? thanks
  5. Dec 22, 2008 #4
    hang on no i havn't read this properly at all, i'm still as stuck as i first was where has little v and u came from? if anything i'm more confused
  6. Dec 22, 2008 #5
    think i'm getting my head arround it now u and v elements of UnW so that u+v are elements of U and u+v are elements of W so because of this u+v are elements of UnW
    the only thing i'm not getting is where this u and v are coming from? It would make more sence to me if it was u and w not u and v. and where is the w surely this is an element of UnW also?????

    i do however understand rules 1 and 3 multiplying by a scalar and zero vector just the addition thats baffeling me
    Last edited: Dec 22, 2008
  7. Dec 22, 2008 #6


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    You can call vectors anything you like! I wanted to avoid "u" and "w" specifically because someone might think that "u" must be in U and "w" must be in W.

    Perhaps it would be better to say "if x and y are members of UnW then they are both members of U and both members of W. Since U is a subspace, it is closed under addition: x+ y is in U. Since W is a subspace, it is closed under addition: x+ y is in W. Since x+ y is in both U and W, it is UnW.

    Now, you need to show that if x is in UnW, so is ax for a any number.

    Oh, and you need to prove the subset is non-empty- most often that's done by showing that the 0 vector is in the set.
  8. Dec 26, 2008 #7
    thanks got it now
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