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Intersection of disjoint SETS is empty

  1. Sep 1, 2004 #1
    The question is:
    Suppose [tex]W[/tex] and [tex]X[/tex] are subspaces of [tex]R^8[/tex].
    Show that if dim[tex]W=3[/tex], dim[tex]X=5[/tex], and [tex]W+X = R^8[/tex], then [tex]W \cap X = \{0\}[/tex].

    I can see this is obvious iff W and X are disjoint sets. If we add members of [tex]W[/tex] to [tex]X[/tex] in the usual way, and we get the new set [tex]W+X[/tex] whose dimension is now 8 (given), then the only way the dimension can be 8 is if each member of [tex]W[/tex] and [tex]X[/tex] is 'different', implying that [tex]W[/tex] and [tex]X[/tex] are disjoint and their intersection is zero.

    Here is how I have tried to prove it...

    1. We are given that [tex]R^8 = W + X[/tex]
    2. I have assumed [tex]{W,X}[/tex] are independent - how would I prove this?
    3. If (1) and (2) hold, then [tex]R^8[/tex] is also the direct sum of the two subspaces, [tex]R^8 = W \oplus X[/tex].
    4. Let [tex]\textbf{v}[/tex] be a vector from [tex]W \cap X[/tex] and consider the equation [tex]\textbf{v} = \textbf{v}[/tex].
    5. The left side of that equation is a member of [tex]W[/tex] and the right side is a linear combination of only one member of [tex]X[/tex]. But the independence of the spaces then implies that [tex]\textbf{v} = \textbf{0}[/tex].
    6. Hence the intersection of [tex]W[/tex] and [tex]X[/tex] is trivial.
  2. jcsd
  3. Sep 1, 2004 #2
    For (2)

    [tex]W,X[/tex] are independent because their sum [tex]W+X[/tex] is an eight member subset of an eight member vector space, namely [tex]R^8[/tex].
  4. Sep 1, 2004 #3

    matt grime

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    You are confusing issues here.

    The intersection of disjoint SETS is empty, not zero. For some reason you seem to focus on the fact that vector spaces are sets, this is neither here nor there in this question, it is the fact that they are vector spaces, and as such the intersection of them is also a vector space, albeit the zero vector space in this case. The y are certainly not disjoint sets, in fact they cannot be disjoint sets.

    You should have proven that WnX is a vector space. You should have proven that

    dim(W+X) = dimW + dimX -dim(WnX)

    from which the answer to your question follows easily.

    W is not an 8 member vector space. Ok, I'm guessing here as I've never heard of such a thing as an "n member" vector space in this context.

    You seem to be confusing a Vector space with its spanning set. They aren't the same thing.

    You can provre all these things, if you've not seen how, by an argument that goes a little like:

    WnX is a subvector space of W and X, let a,b,..,c be a basis of it, complete to two bases, one of X and one of W a,b,..,c,d,..,e and a,b,c,f,..,g respectively. The combined sets clearly span, and are also clearly linearly independent. (Note set here refers to the set of vectors in the basis, not the vector space). By counting we see that

    dim(W+X) = dimW + dimX -dim(WnX)

    your answer is now a corollary.
  5. Sep 1, 2004 #4
    Thanks Matt, I knew that would be the way to do it.

    One query though...

    ..what exactly are you doing here?
  6. Sep 2, 2004 #5

    matt grime

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    The standard way of working with bases.

    we know WnX < X < V
    and WnX<W<V

    you may choose a basis of a subspace, then, by adding more vectors get a basis of the containing space. I was just writing that out. Would it be clearer if I wrote

    let [tex]u_1, u_2, \ldots, u_s[/tex] be a basis of WnX, by the standard theorems of (finite) dimensional vector spaces you may add more vectors to it [tex]x_{s+1},\ldots,x_r[/tex] to get a basis of X, and vectors [tex]w_{s+1},\ldots,w_{t}[/tex] to get a basis of W

    then by counting t+r-s is the dimension of W+X
  7. Sep 2, 2004 #6

    matt grime

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    It occurs to me that you might not have seen some of the ideas I'm assuming you have. So here's a recap.

    We know what a vector space is, take that as read, generically denote a vector space as V.

    Given a set of vectors in V, S={s_1,s_2,.., s_r} we can form a vector space called the span of S, and written <S>. It is the smallest vector subspace of V containing all the elements in S. We say that S spans V if <S>=V.

    Given any set of vectors, S as above, Sylvester's law of replacement allows us to find anothe set of linearly independent vectors S', which have the same span as S. A linearly independent spanning set is a basis.

    Given a vector space V and a subspace W, there is the quotient space V/W, whose elements are equivalence classes of vectors in V. The relation is x~y iff x-y lies in W.

    So, given W<V and a basis of W we can complete to a basis of V by the following method:
    Let x_1,..,x_r be a basis of W
    we pick some basis of V/W, say [x_{r+1}],..., [x_{n}] recalling that elements in the quotient space are equivalence classes of elements in V, and we use [y] to denote the equivalence class of y.

    then the set x_1,..,x_n is a basis of of V.
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