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Prove the sum of two subspaces is also a subspace.

  1. Sep 6, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi I'm trying to prove that the sum of two subspaces [tex]U[/tex] and [tex]W[/tex] is also a subspace.

    2. Relevant equations

    [tex]U[/tex] is a subspace of [tex]V[/tex] if [tex]U[/tex] is also a vector space and it contains the additive identity, is closed under addition, and closed under scalar multiplication.

    The definition of a sum a vector subspace U and W is
    [tex]U+W=\{u+w:\, u\in U,w\in W\}[/tex]

    3. The attempt at a solution

    1. Since [tex]U[/tex] and [tex]W[/tex] both contain the additive identity, [tex]U+W[/tex] contains the additive identity



    3. Since both [tex]U[/tex] and [tex]W[/tex] are closed under scalar multiplication, any combination of [tex]u+w[/tex] is closed under scalar multiplication since multiplication is distributive, associative and commutes (assuming were dealing with the reals here).


    I'm having a hard time thinking about how to justify that U+W is closed under addition.

    Also is my justification for closure under scalar multiplication right?
     
    Last edited: Sep 7, 2009
  2. jcsd
  3. Sep 7, 2009 #2

    lanedance

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    Homework Helper

    First let Z = U+W

    to show its closed under addition
    so take z, z' in Z then show z + z' is an element of Z

    your words are close, but I would try a similar thiong for the scalar multiplication to be a little more explicit
     
  4. Sep 7, 2009 #3

    VietDao29

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    Homework Helper

    [tex]U+W=\{u+w:\, u\in U,w\in W\}[/tex]

    Writing this means that: U + W is the set of the sum of all possible 2 vectors, of which one is taken out from U, and the other from W, i.e:
    • If u is some vector of U, and w is some vector of W, then [itex]u + w \in U + W[/itex]. (1)
    • And if v is some vector of U + W, then it can be split into the sum of 2 vectors, one in U, and one in W. That is: [tex]v \in U + W \Rightarrow \exists u \in U, w \in W : v = u + w[/itex] (2).

    Ok, this is good, but you really need to show why it contains the additive identity, instead of just reasoning like that. You can go like this:

    Since U, and W are subspaces, that means:

    [tex]\left\{ \begin{array}{c} 0 \in U \\ 0 \in W \end{array} \right. \Rightarrow 0 = 0 + 0 \in U + V[/tex] (due to (1)) (the underline part is just for you to understand why)

    As I said above, you should make this part clearer.

    Just start out as normal. So, what you want is to prove that:

    [tex]\forall v_1 \in U + W, v_2 \in U + W[/tex], we must have [tex]v_1 + v_2 \in U + W[/tex]

    Hint: Use (2), and the fact that both U, and W are already subspaces. :)

    The same proof goes for "closed under scalar multiplication".
     
  5. Sep 7, 2009 #4
    Ok thank you. Let me think about this some and try to make some progress. I'll be back later in the day.
     
  6. Sep 7, 2009 #5
    OK I think I got it. Thanks guys!!
     
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