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Subspace proof help

  1. Mar 18, 2009 #1
    I saw this problem in a book, it asks if there are two subspaces of Rn, say U & V and the following condition is true:
    W={[tex]w \in R^n[/tex] : w=u+v for some [tex]u \in U [/tex] and [tex]v \in V[/tex]}

    Make a proof/show that W is a subspace of Rn.

    I think maybe we need to try to somehow prove that the set W is a subspace of Rn by showing that it's non-empty and closed under addition/scalar multipication. Does anyone know how to show this? I'm not sure how we can do it, any explanation or links would be appreciated.

    Thanks.
     
  2. jcsd
  3. Mar 18, 2009 #2

    HallsofIvy

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    Re: Subspaces

    I don't see why just straightforward verification of the conditions you state wouldn't work.

    W is not empty because U and V both contain the 0 vector. So what vector is in W?

    Suppose w and w' are in W, we can write w= u+ v and w'= u'+ v' where u and u' are in U, v and v' are in V. what can you say about w+ w'? What about [itex]\alpha w[/itex] for [itex]\alpha[/itex] a number?
     
  4. Mar 18, 2009 #3
    Re: Subspaces

    w+ w' = (u+v)+(u'+v'), therefore it's closed under addition. And for some [tex]\alpha \in R[/tex], [tex]\alpha w= \alpha u + \alpha v[/tex] & [tex]\alpha (w + w')= \alpha (u + v) + \alpha (u'+v')[/tex] so closed under scalar multipication. Is this right?

    Could you please explain because I'm not sure. How do we write a proof to show that U and V contain the 0 vector?
     
  5. Mar 18, 2009 #4

    HallsofIvy

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    Re: Subspaces

    Every subspace contains the 0 vector! Every subspace is non empty so either it contains the 0 vector or it contains some non-zero vector v. Since a subspace is closed under scalar multiplication, it contains (-1)v= -v. Since a subspace is closed under addition, it contains v+ (-v)= 0.

    If you don't want to use 0 specifically, you can argue that U is a subspace so contains some vector u, V is a subspace so contains some vector v, therefore U+ V contains u+v and so is non-empty.
     
  6. Mar 18, 2009 #5
    Re: Subspaces

    It is right. This is because, for any [tex]\alpha \in R[/tex], [tex]\alpha w= \alpha u + \alpha v\in {R^n}[/tex]. why? recall that u and v are vectors respectively in subspaces U and V. and also for any w in W, w = u + v implies u is in U and v is in V.But
    [tex]\alpha u[/tex] is in U and [tex]\alpha v[/tex] is V. Now think of the sum, their sum will surely be in W.

    U and V contains the zero vector because of the fact that a linear space is an additive abelian group. You recall that a group must contain the identity element. Thus for a linear space, the identity element is the zero vector under the binary operation '+'. you recall that a subspace of a linear space is linear space in its own right. Thus, the subspaces U and V must contain the zero vector.
     
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