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Vector Spaces and Dimensions

  1. Apr 23, 2012 #1
    1. The problem statement, all variables and given/known data



    2. Relevant equations
    From my notes I'm aware of the following equation: dim(A + B) = dimA + dimB − dim(A ∩ B).


    3. The attempt at a solution
    I'm assuming part of the solution involves the equation above and rearranging it but I'm not sure how I would determine dim(A + B). I also know A + B := <A ∪ B> = = {a + b: a ∈ A, b ∈ B} but I can't see where to go.

    Thanks in advance!
     
    Last edited: Apr 23, 2012
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  3. Apr 23, 2012 #2

    HallsofIvy

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    Since dim(B)> dim(A), it is not possible for B to be a subset of A but consider the situation of A being a subset of B. That will give the largest possible size of [itex]A\subset B[/itex]. To find the smallest possible size think about how separate A and B can be. For example if U had only dimension 9, all of A would have to be in B! But if U had dimension 17= 8+ 9, A and be could be "completely" separate, having only the 0 vector in common.
     
  4. Apr 23, 2012 #3
    I'm not too sure where to go from what you said.

    Following from having only the 0 vector in common this would mean A and B would only contain one element and so have dimension 1?

    Also is the equation I stated relevant because then would I also need to work out A + B and thus dim(A + B).

    I'm sure what you said probably is a big hint but I can't seem to figure out where to go, could you please provide more help?

    Thanks
     
  5. Apr 23, 2012 #4

    HallsofIvy

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    No, that was not my point! A and B cannot have "only the zero vector in common" since then the dimension of their direct sum would be 8+ 9= 17 which is impossible in only 10 dimensions. That's the key idea! As you say, [itex]dim(A+ B)= dim(A)+ dim(B)- dim(A\cap B)[/itex]. Here dim(A)+ dim(B)= 8+ 9= 17 while dim(A+ B) cannot be larger than 10. So, at most, [itex]10= 8+ 9- dim(A\cap B)[/itex].

    (The dimension of the "space" containing only the zero vector is 0, not 1, by the way.)
     
  6. Apr 23, 2012 #5
    Oh right I see. So dim(A∩B) must be at least 7 and at most 17?
     
  7. Apr 23, 2012 #6

    HallsofIvy

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    No, that's not at all right! [itex]A\cap B[/itex] is a subset of A so it can't possibly have dimension greater than the dimension of A.
     
  8. Apr 23, 2012 #7
    Apologies for causing any frustration but after having a rethink I have now arrived at the following(which could be wrong again),

    A+B is a subset of V, therefore,
    dim(A+B) <= dim(V) = 10, therefore,
    dim(A+B) <= 10.

    Then, dim(A+B) = dim(A) + dim(B) - dim(A∩B) = 17-dim(A∩B) <= 10
    and so, dim(A∩B) >= 7.

    But A∩B is a subset of A and a subset of B, thus
    dim(A∩B) <= dim(A) = 8
    and dim(A∩B) <= dim(B) =9.

    Therefore, dim(A∩B) <= 8.

    Giving, dim(A∩B) = 7 or 8?
     
  9. Apr 23, 2012 #8
    Could someone check what I wrote above please?

    Thanks.
     
  10. Apr 23, 2012 #9

    micromass

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    What you wrote is correct.

    But please wait before bumping your thread after 24 hours have passed, thank you.
     
  11. Apr 23, 2012 #10
    Okay thanks and apologies.

    Could anyone help me out with the part concerning "Give an example illustrating each
    possible value." What are they actually asking for?
     
  12. Apr 23, 2012 #11

    micromass

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    They want you to give specific examples of V, A, B such that dim(A)=8, dim(B)=9 and [itex]dim(A\cap B)[/itex] is 8 or 9.

    I suggest taking [itex]V=\mathbb{R}^{10}[/itex], then dim(V)=10 as required. Now, how would you pick A and B?
     
  13. Apr 23, 2012 #12
    I'm not totally sure but could you take A = {(a,b,c,d,e,f,g,h,0,0) | a,b,c,d,e,f,g,h are Real} and B = {(a,b,c,d,e,f,g,h,i,0) |a,b,c,d,e,f,g,h,i are Real}?

    Somehow I think that's wrong though :s
     
  14. Apr 23, 2012 #13

    micromass

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    OK, that's good. What is [itex]dim(A\cap B)[/itex] in this case?
     
  15. Apr 23, 2012 #14
    Would it equal to dim(A) and so 8?
     
  16. Apr 23, 2012 #15

    micromass

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    Correct!! So that's one example.

    Now, can you find an example where [itex]dim(A\cap B)=7[/itex]??
    Hint: A and B have to span V.
     
  17. Apr 23, 2012 #16
    Does it not need to equal 7? As my answer before was dim(A∩B) = 7 or 8, not 8 or 9?
     
  18. Apr 23, 2012 #17

    micromass

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    Yes, sorry.
     
  19. Apr 23, 2012 #18
    For the answer of 7 would the following be okay?

    A={(a,b,0,c,d,e,f,g,0,h) |a,b,c,d,e,f,g,h are Real}
    B={(a,b,0,c,d,e,f,g,h,i) |a,b,c,d,e,f,g,h,i are Real}
    Then dim A = 8, dim B = 9,
    but A∩B = ({a,b,0,c,d,e,f,g,0,0)|a,b,c,d,e,f,g are Real} and this has dimension 7?

    Also going back to my earlier working was it correct to say A∩B was a subset of A of should I have said it is a subspace of A?

    Thanks for all the help so far it much appreciated.
     
  20. Apr 23, 2012 #19

    micromass

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    This is incorrect. The correct intersection is

    [tex]A\cap B=\{(a,b,0,c,d,e,f,g,0,h)|a,b,c,d,e,f,g,h\in \mathbb{R}\}[/tex]

    (notice the h in the last coordinate which is nonzero unlike your example).

    Like I said, you need to make sure that A and B together span the space.

    It's both a subset and a subspace, so saying it's a subset isn't wrong. But it indeed seems more appropriate to call it a subspace here.
     
  21. Apr 23, 2012 #20
    I'm not getting very far, could you give another hint perhaps?

    Thanks.
     
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