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Dimension of intersection of U and V

  1. Feb 22, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove true:
    For any subspaces U,V of R^n dim(U intersect V) <= min(dim(U), dim(V))

    2. Relevant equations
    Min(a,b) = the minimum value of A and B

    3. The attempt at a solution
    I know this statement is true however I can't quite figure out where to start on how to prove this...Help please.
  2. jcsd
  3. Feb 22, 2010 #2
    The best way to approach this is to think about the subspace generated by intersection(U,V)... call it W for short. The dimension refers to the size of the basis, so any vector on W will also be found on U and found on V. If you think about V being the bigger one (say size m for dim|V| and size k for dim|U|), a vector w on W will have an expression as a linear combination in the basis of V and in the basis of U. So w can be written as a combination of of m vectors on V , and k vectors on U (k<m). Since they are subspace of the same vector space, they get their basis from the same vector space. Since W is also a subspace of the larger vector space, it gets its basis from that too.

    Think of it in the sense that , in order for a vector to be on the basis of W, it has to be on the basis of V and U, since U has less vectors in its basis, there is going to be at least (m-k) vectors on the basis of V that will not be on the basis of W.
  4. Feb 22, 2010 #3
    Ok you have quite a few typos in your statement so what you are trying to say at some points isn't quite clear.
  5. Feb 22, 2010 #4


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    (U intersect V) is contained in U and V. Try that.
  6. Feb 22, 2010 #5
    Well I'd say use lagrange's formula from group theory, but I'm not sure how to delineate that.
    OK say you have the original vector space X, where X has the basis {e1,e2,e3...en}
    for a subspace V of X, the basis of V is a subcollection of the basis of X , ie V has a basis {e1,e5,e7,..e(n-1)} say for a grand total of M vectors in the basis.

    now consider another subspace U of X, but this one has {e1,e3,e8,...en} , we'll say K vectors, where K<M.

    Now consider the basis of the intersection of U and V =W.
    If every vector in the basis of U is in W, then W has dimension k. V will have extra basis vectors that U will not have, so Dim(u,v) would be k. where k is the min (dim(U),dim(V))
    but thats assuming that every vector on the basis of U is found on the basis of V.
    So that means the dim(u,v) would be less than the min(dim(u),dim(v))
    like say X has basis {e1,e2,e3,e4,e5,e6}Dim(X)=6 and U has {e1,e4,e5} (dim U=3) and V has {e1,e3,e4,e6} dim(v)=4
    then the only vectors on the basis of W would be e1 and e3 , so it'd have dim(2)
  7. Feb 22, 2010 #6
    Thank you for editing your statement.
    I proved it using the properties of a p dimensional subspace which actually works a lot better and is simpler than what you are trying to say.
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