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Showing that a set of linear algebra statements are true

  1. Mar 11, 2013 #1
    V is the direct sum of W1 and W2 in symbols:

    V = W1 (+) W2 if:

    V = W1 + W2 and
    W1 [itex]\cap[/itex] W2 = {0}

    Show that the following statements are equivalent:

    1. V = W1 (+) W2

    2. Every vector v[itex]\in[/itex]V can be written uniquely as w1 + w2 where w1[itex]\in[/itex]W1 and w2[itex]\in[/itex]W2

    3. V = W1 + W2 and for vectors w1[itex]\in[/itex]W1 and w2[itex]\in[/itex]W2 if w1 +w2 = 0 then w1 = w2

    4. If [itex]\alpha[/itex]1 is a basis for W1 and [itex]\alpha[/itex]2 is a basis for W2 then
    [itex]\alpha[/itex] = [itex]\alpha[/itex]1 [itex]\cup[/itex] [itex]\alpha[/itex]2

    Attempt:

    I'm thrown off here because I'm given the definition of a direct sum. So how can I even show that all these statements are equivalent? Would I start by assuming one of the statments is true? For example: Let's say statement 1 is true. that means......well it means exactly what statement 2 is. Maybe take a vector in W1 and one in W2 sum them together and create a vector in V? But then to do that I would have to define an exact vector space.......
     
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  3. Mar 11, 2013 #2

    Mark44

    Staff: Mentor

    Start by assuming that statement 1 is true. Then show that this implies that statement 2 must be true, which you can do by invoking the definition of a direct sum.

    Next show that statement 2 being true implies that statement 3 is true. Then show that statement 3 being true implies that statement 4 is true. Finally, show that statement 4 implies statement 1.

    You don't have to define a vector space. You don't mention it, but it seems to me that V is some arbitrary vector space and that W1 and W2 are subspaces of V.
     
  4. Mar 11, 2013 #3

    But then what is it that I'm really doing beyond repeating the definition of a direct sum? Because it seems that those each just logically follow each other. Yes it is suppose to be an arbitrary vector space.
     
  5. Mar 11, 2013 #4

    Dick

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    Yes, they do follow from each other. That's exactly what you are supposed to show. Each statement is different from the definition of direct sum. You have to show why they are equivalent.
     
  6. Mar 12, 2013 #5
    Ok, I'm still puzzled at how I can show it. Because if I didn't have the 4 statements for example. Starting with the definition of a direct sum, I would've just said that the definintion of a direct sum implies that every v[itex]\in[/itex] V can be written uniquely as a combination of w1 and w2. I don't see how I'm "showing" this though.
     
  7. Mar 12, 2013 #6

    Dick

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    Show HOW the implication works. You start with v=w1+w2=w1'+w2'. What property of the direct sum let's you say w1=w1' and w2=w2'? This isn't a hard proof by any means, but there is something to say.
     
  8. Mar 12, 2013 #7


    Well the only properties I have are that the intersection of the two sets is {0}. Also each vector in the subspace is unique. I suppose by uniqueness:

    there exists a unique vector in w1 and w2 s.t v is the combination of them?.......Still not seeing it.
     
  9. Mar 12, 2013 #8

    Dick

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    You don't HAVE uniqueness, you want to SHOW uniqueness. If v=w1+w2=w1'+w2' (w1, w1' in W1 and w2, w2' in W2) then w1-w1'=w2'-w2. So?
     
  10. Mar 12, 2013 #9

    I'm going to go out on a limb and say 0=0, but I don't think that would make sense considering I'm trying to show uniqueness so I can't assume w1-w1' = 0 and the same for w2. And so my struggle continues with this sort of thing being my impediment from obtaining A's. sigh.
     
  11. Mar 12, 2013 #10

    Dick

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    Which subspace is w1-w1' in? Same question for w2-w2'.
     
  12. Mar 12, 2013 #11
    w1-w1' is in W1 and w2-w2' is in W2. ==> W1 = W2 but that doesn't do anything.

    unless:

    v = (w1-w1') + (w2 - w2')
     
  13. Mar 12, 2013 #12

    Dick

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    So call u=w1-w1'=w2'-w2. From what you have said that must mean u is in W1 and W2. What's the only vector that's both in W1 and W2?
     
  14. Mar 12, 2013 #13
    the only vector in W1 and W2 is the 0 vector. I'm still puzzled as to how this shows statement 2.
     
  15. Mar 12, 2013 #14

    Dick

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    So u=0. Then w1-w1'=0 and w2'-w2=0. Remember what you want to prove?
     
  16. Mar 12, 2013 #15
    As far as I remember I'm trying to prove statement 2. Then with that I'll prove statement 3, etc. But the fact that u = 0, how is this showing statement 2? I'm trying to go over the steps but I'm failing to see the connection
     
  17. Mar 12, 2013 #16

    Dick

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    Don't worry about the parts yet. You are trying to show 1)->2). You want to show that if v=w1+w2=w1'+w2', then w1=w1' and w2=w2'. I think you are practically there is if you string all the stuff we've been doing together.
     
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