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Finding Dimensions

  1. Oct 7, 2012 #1
    I've attached the problem.

    For S:
    I form the matrix:
    1 0
    0 1
    0 0
    0 0

    Thus the dimension is 2.

    For T:
    I form the matrix
    0 0
    1 0
    0 1
    0 0
    Thus the dimension is also 2.

    Is that the correct idea?



    Also what does S ∩ T mean? I couldn't find the symbol in my textbook.
     

    Attached Files:

  2. jcsd
  3. Oct 8, 2012 #2
    Yes, the dimension for those matrices are 2, when talking about dimension we either talk about row or column dimension.
    the S ∩ T means all the elements that are common in S and T.
     
  4. Oct 8, 2012 #3
    What does common mean? A column that is included in both matrices?
     
  5. Oct 8, 2012 #4
    Well in this case it would be the rows that both S and T have common elements in.
    Hint: look past the parameters s,t
     
  6. Oct 8, 2012 #5
    Hmmm I don't understand why in this case it would be rows?

    The only row in common is the last row (0 0).
     
  7. Oct 9, 2012 #6
    Sorry I misread your problem, since there are no common elements in each vector, there is no dimension for the intersection of S and T because they have no common elements
     
  8. Oct 9, 2012 #7

    HallsofIvy

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    First "[itex]\cap[/itex]" is the standard symbol for the "intersection" of two sets- it is the set contain all things that are in both sets. I am surprised that you are working with vector spaces but have not yet learned basic set terminology.

    Second, this problem does NOT ask you to find the "intersection" of two vectors- vectors are not sets. It asks you to find the subspaces spanned by the two vectors. Any vector in S is of the form <a, 0, 0, 0> so this is a one dimensional space. In particular, it has second, third and fourth components 0. Any vector in T is of the form <0, b, 0, 0>. In particular, it has first, third and fourth components 0. Any vector in their intersection must be in both and so must sastisfy both conditions. The only vector that does that is <0, 0, 0, 0>, the zero vector. That is a subspace of dimension 0.
     
  9. Oct 9, 2012 #8
    Aren't vectors in S of the form <s 0 0 0>and <0 t 0 0>
    and vectors in T of the form <0 s 0 0> and <0 0 t 0>?
     
  10. Oct 9, 2012 #9

    HallsofIvy

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    Yes, S is spanned by <1, 0, 0, 0> and <0, 1, 0, 0> and T is spanned by <0, 1, 0, 0> and <0, 0, 1, 0>. And since those are independent, they are bases for the subspaces so each has dimension 2.

    The fact that <0, 1, 0, 0> is in both of those should make this problem of the intersection easy!
     
  11. Oct 9, 2012 #10
    Oh so the dimension of S∩T is just 1.
     
  12. Oct 11, 2012 #11
    For this case, when you say that the dimension of S∩T is one, would it be necessary to list the vector 0 1 0 0 as the one that exists in both? I guess I'm not sure how to to state it. I can obviously just say dim(S∩T)=1, but do I need to explain it in words or justify it?
     
  13. Oct 11, 2012 #12

    HallsofIvy

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    Typically, with a problem of this kind you should justify your answer. Any vector in S is of the form a(1, 0, 0, 0)+ b(0, 1, 0, 0) and any vector in T is of the form c(0, 1, 0, 0)+ d(0, 0, 1, 0). Any vector that is in their intersection can be written both ways so we must have a(1, 0, 0, 0)+ b(0, 1, 0, 0)= c(0, 1, 0, 0)+ d(0, 0, 1, 0) for some numbers a, b, c, and d. That is the same as (a, b, 0, 0)= (0, c, d, 0) so we must have a= 0, b= c, d= 0.

    (Be very careful with your wording. You cannot talk about "the vector that exists in both" because the intersection, being a subspace, contains an infinite number of vectors.)
     
  14. Oct 11, 2012 #13
    Yeah, I just talked to my professor about it and he said the dimension is the size of a basis. By "size of a basis," does that just mean how many vectors span S∩T, which is likewise the dimension of S∩T. If so the basis of S∩T is spanned by the vector [ 0 1 0 0]^t. Thus the size of the basis is 1 and the dimension is 1.

    Is that the right idea?
     
  15. Oct 11, 2012 #14
    oops
     
    Last edited: Oct 12, 2012
  16. Oct 11, 2012 #15
    I think you got it. [0 1 0 0]^t span both S and T, thus dim(SnT)=1.

    I could be wrong though. Anyone else verify?
     
  17. Oct 11, 2012 #16
    Hmmm. There's a relationship dim(U+V)=dim(U)+dim(V)-dim(U ∩ T)
    Where U is your S and V is your T.
    Dim(U+V)=2=2+2-dim(U ∩ T)
    Which makes dim(U ∩ T)=2
     
  18. Oct 12, 2012 #17
    hmmm That equation is saying there's 2, but i think hallsofivy says there's 1. I'm really confused. That equation does make sense tho, U+V is spanned by 2 linearly independent vectors.
     
  19. Oct 13, 2012 #18

    HallsofIvy

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    No, U+ V (Your S+ T) is spanned by three linearly independent vectors {< 1, 0, 0, 0>, <0, 1, 0, 0>, and <0, 0, 1, 0>, the three vectors you gave initially. So dim(U+ V)= 3, not two.
    so dim(U+ V)= 3= 2+ 2- dim(U∩V)= 2+ 2- 1.
     
  20. Oct 13, 2012 #19
    Oh I think I misunderstood the meaning of dim(U+V). I thought you add U+V then find the dimension.
     
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