Prove span(S1 U S2) = span(S1) + span(S2)

  1. Show that if S1 and S2 are arbitrary subsets of a vector space V, then span(S1 U S2) = span(S1) + span(S2)

    When attempting it i used the definition of a union as well as the definition of the sum of two sets and I also used the fact that the union of two sets is the sum of all the elements in set1 plus all elements in set2 minus the intersection of the set. By this method, the only way to prove the above statement is for the intersection of the sets to be equal to zero, and it seems pretty obvious to me that the intersection of any two sets in V does not necessarily have to equal 0. Unless the word "abitrary" implies that? So just wondering if anyone can help me with this.
     
  2. jcsd
  3. matt grime

    matt grime 9,396
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    When showing two sets are equal always first (try to) show that something in one set is in the other and vice versa. In this case this gives the solution very quickly. In fact so easily that there really is nothing to prove; it's one of those questions that is so easy that its proof seems unobvious.I mean, what is the span of something? it is the set of all possible sums of elements in it, so the sum of possible sums is the possible sums of sums.... ok, that appears confusing but it isn't, i'm just saying that (x+y)+(u+v) = (x+y+u+v)

    You can forget the union minus intersection part, it is completely irrelevant.
     
    Last edited: Nov 25, 2005
  4. Ok but what if I do it my way, using the definitions:

    span(S1) + span(S2)
    = {sum from i=1 to n[a_i * x_1i]: ai E F, x1i E S1 } + {sum from j=1 to n[ai * x2i]: ai E F, x2i E S2 }
    = { s1 + s2: s1 E span(S1), s2 E span(S2) }
    = { sum from i to n[ai * x1i] + sum from i to n[aj * x1j] }

    span(S1 U S2)
    = span({x: x E S1 or x E S2})
    = { sum from k to p[ak * x]: x E S1 or x E S2 }
    = { sum from i to n[ai * x1i] + sum from j to m[aj * x2j] - sum from l to q[al * x3l]: x3 E S1 and x3 E S2 }

    Note: * here means multiplication, E means 'element'

    See if I put it in that form, it seems to be unequal. What am doing wrong here? Thanks for the reply btw.
     
  5. matt grime

    matt grime 9,396
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    That seems horribly notationally overloaded. And just because things aren't obviously the same because of two different descriptions means nothing.

    Just show set equality. Some thing is in span(U_1)+span(U_2) if it is of the form a+b for a in span(U_1) and b in span(U_2) which is obviously something in span(U_1uU_2)

    The other containment is as hard to demonstrate: split something in the span of (U_1uU_2) into the sum of things in U_1 and things in U_2
     
  6. just use the definitions

    just use the definition.

    let K be from Span(S1 U S2)
    then K = a1V1 + ... + anVn ... + b1U1 + ... + bmUm +...
    where Vi's are from S1 and Ui's from S2 for some ai's and bi's from the field
    now all you have to do is group all the Vi's and group all the Ui's
    K = (a1V1 + ... + anVn ... ) + (b1U1 + ... + bmUm +... )
    and it is clearly seen that
    K = W1 + X1 where W1 is from Span(S1) and X1 is from Span(S2)
    therefore K belongs in Span(S1) + Span(S2)

    now for the reverse just go exactly backwards providing necessary explanations.

    Was there a point that was difficult to see?
     
  7. Ya, i already figured it out. I guess the problem i made was assuming that just because some vectors are added twice in the span(s1) + span(s2) and only one of them is added in span(s1Us2) that the results are different, but its not different, because the linear combinations include an infinite amount of scalars from a field and having 2*a*v1 in span(s1)+span(s2) and a*v1 in span(s1 U s2) is the same because the a E of field can have an infinite amount of values.
    Or is my assumption that a field contains an infinite amount of elements incorrect?
     
  8. matt grime

    matt grime 9,396
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    a field does not have to contain an infinte number of elements. nor do i see what that has to do with your argument at all.
     
  9. well, the span of a set of vectors is the set of all possible linear combinations them, i.e a1v1+...+anvn where each ai is an element of a field. Now, spanS1 + spanS2 is the set of all linear combinations in S1 + linear combinations in S2. span(s1 U s2) is the same thing but whenever a vector is found in both S1 and S2, instead of being added with itself again, it is just added once. So whenever a vector "vb" is found in both, spanS1 + spanS2 includes 2*(vb)*(element of a field) for all possible elements of a field, and span(S1 U S2) includes only (vb)*(element of a field) for all possible elements of a field. Now i dont know the exact properties of a field, but i assume if a is an element of a field then a + a is also an element of a field, like in the real numbers. Cause if that was the case it wouldnt matter if you have 2 equal vectors in both sets but just varying by a multiplicative factor of 2, because if you look further youll be able to find by manipulating "a" one that matches with the one that's off by *2, and of course you can do an infinite amount of these manipulations..
     
  10. matt grime

    matt grime 9,396
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  11. ya youre right, i realized its possible to have a field with a finite amount of elements. Like the field with only 0 and 1 and 1 + 1 = 0. But still, my argument holds, because for every 2 equal vectors summing up to a*vb (in span1 + span2) you can always find a*vb in span(1U2) by manipulating the a. (Kind of like what you said in the beginning about showing that something in one set is in the other and vice versa).
     
  12. matt grime

    matt grime 9,396
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    i think you need to learn to typeset more clearly, at least try to indicate subscripts and superscipts, and don't use vb to represent something other than v times b where v and b are other things.
     
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