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Proving a subspace

  1. Aug 10, 2011 #1
    1. The problem statement, all variables and given/known data

    Let U and W be subspaces of a vector space V.

    Show that the set

    U + W = {v (element symbol) V : v = u + w, where u (element symbol) U and w (element symbol) W}

    Is a subspace of V.

    2. Relevant equations

    -

    3. The attempt at a solution

    I really don't know where to start here? What would I work with? How can I add these vectors to show that their sum is in fact forming a subspace of V? I just don't know where to start?
     
  2. jcsd
  3. Aug 10, 2011 #2
    This is equivalent to showing that V is closed under addition. So, what's the definition of V, and how is the addition operator defined?
     
  4. Aug 10, 2011 #3
    Well, we are given that V is a vector space, therefore, by our axiom (1), if u and w are in V, then u + w is in V. But we are not given that u and w are in V, we are given that u and w are in U and W, which are both subspaces of V.

    So, is it possible to show that u + w is a subspace of V?
     
  5. Aug 10, 2011 #4
    If [itex]U \subset V[/itex] and [itex]u \in U[/itex], is [itex]u \in V[/itex]?
     
  6. Aug 10, 2011 #5
    I'm not sure what you are asking? Sorry...
     
  7. Aug 10, 2011 #6
    If u is an element in the space U, and if U is a subspace of V, does this imply that u is an element of V? What is the definition of a subspace?

    Answering that should tell you how to answer the question you asked in #3.
     
  8. Aug 10, 2011 #7
    A subspace of a vector space V is a subset of V which itself is a vector space under the addition and scalar multiplication defined on V.

    Ok, this makes sense, I suppose I just was not looking at it properly.

    So this kind of proof, it would mainly be in words as I can imagine it. I would write:

    u and w are both elements of V, since they are elements of subspaces of V. And, by axiom (1), which states that if u and w are elements of V, then u + w is also an element of V. Thus, since u and w are elements of a subspace (which are of course vector spaces), u + w is itself a subspace of V.

    ??? Is this okay?

    Thanks.
     
  9. Aug 10, 2011 #8
    That's the general logic. The last step (proving that u+w is a valid subspace) may or may not need more rigor depending on the assumptions of the problem.
     
  10. Aug 10, 2011 #9

    vela

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    U+W and u+w are not the same thing. u+w is an element of V; it is not a subspace of V. U+W, on the other hand, is what you're trying to prove is a subspace.
    No, it's not. You're trying to prove that U+W is a subspace. That means you need to show that if [itex]x_1, x_2 \in U+W[/itex], then

    1. [itex]x_1+x_2 \in U+W[/itex];

    2. [itex]c x_1 \in U+W[/itex], where c is a constant; and

    3. [itex]0 \in U+W[/itex]

    Note that you're trying to show that the various elements are in U+W, not V. You're proving U+W is non-empty and is closed under addition and scalar multiplication.
     
  11. Aug 10, 2011 #10
    But it is the first step. As long as he understands the logic behind the construction of U+W (which is where it seems he was stuck) then proving that U+W satisfies the properties of a subspace is relatively easy.
     
  12. Aug 10, 2011 #11

    vela

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    I'll have to disagree. Showing V is closed under addition is axiomatically true since V is a vector space, and the OP seemed to be under the impression that it was all he needed to show to prove U+W is a subspace. It misses the whole point that you have to show it is U+W, not V, that is closed under both addition and scalar multiplication.
     
  13. Aug 15, 2011 #12
    Ok you see. This is my problem. I am not given any constructs of what U and W are, just that they are both subspaces of V. How do I go about showing that if they are both subspaces, then U+W is itself a subspace of V? Thinking about it, I can only conclude that my worded answer is correct?
     
  14. Aug 15, 2011 #13

    vela

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    You don't need to know that level of detail for U and W. You only need to know they are subspaces.

    Let [itex]x_1 \in U+W[/itex]. By definition of U+W, you can say [itex]x_1 = u_1 + w_1[/itex] where [itex]u_1 \in U[/itex] and [itex]w_1 \in W[/itex]. You can say the same sort of thing for [itex]x_2 \in U+W[/itex]. Now look at the sum
    [tex]x = x_1 + x_2 = (u_1 + w_1) + (u_2+w_2) = (u_1+u_2) + (w_1+w_2)[/tex]
    You want to show that this vector x is in U+W. In other words, prove it's equal to the sum of two vectors u and w where [itex]u \in U[/itex] and [itex]w \in W[/itex]. How can you do this using the fact that U and W are subspaces of V?
     
  15. Aug 15, 2011 #14
    Ok. I would imagine that would involve using the fact that we are given that

    velement of V = u + w, where uelement of U and welement of W...??

    Using this piece of information, I could say that:

    By definition of U+W, u = u1 + w1, where u1 is an element of U and w1 is an element of W. And w = w2 + u2 where w2 is an element of W and u2 is an element of U.

    Now, v = u + w
    = (u1 + w1) + (w2 + u2
    = (u1+u2) + (w1 + w2)
    = u + w

    ...look, I don't know...just what exactly am I doing here?
     
  16. Aug 15, 2011 #15

    vela

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    Nooooo...

    You're trying to show x1, x2 ∈ U+W implies that x1+x2 ∈ U+W.

    When you say x1 ∈ U+W, what does that mean given the definition of U+W?
     
  17. Aug 15, 2011 #16
    It means that x1 = u1 + w1, where u1 is an element of U and w1 is an element of W.
     
  18. Aug 15, 2011 #17

    vela

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    OK. You can say the analogous thing for x2. So now you form their sum x = x1+x2. What do you need to show to prove that x is in U+W?
     
  19. Aug 16, 2011 #18
    I would need to show that u1 + w1 + u2 + w2 is in U+W... But I'm not sure how I would do this?
     
  20. Aug 16, 2011 #19

    vela

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    Use the definition of U+W. In other words, you have to show that x is the sum of two vectors, one of which is in U and one of which is in W.

    Hint: To do this, you'll need to rely on the fact that U and W are subspaces so that they are closed under addition.
     
    Last edited: Aug 16, 2011
  21. Aug 16, 2011 #20
    In other words, use axiom (1), which states that if u and v are both elements of a vector space/subspace, then u+v is itself in that space or subspace...correct?
     
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