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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.

  2. jcsd
  3. Mar 18, 2009 #2


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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


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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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