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Proving something is a subset of another set in linear algebra

  1. Nov 22, 2011 #1
    1. The problem statement, all variables and given/known data

    http://people.math.carleton.ca/~mezo/A8math1102-11.pdf

    1b) please

    1. Suppose F is a field, A ∈ Mmn(F), b ∈ Fm and v ∈ Fn is a particular solution to the equation Ax = b. Let S0  ⊆ Fn be the solution set to
    the (homogeneous) equation Ax = 0, and S ∈ Fn be the solution set
    to Ax = b.
    (a) Prove that the set v + S0 = {v + w : w ∈ S0} is a subset of S.
    (b) Prove that S is a subset of v + S0. (Hint: Suppose w ∈ S. Show that w - v ∈ S0. What does this say about w?)

    2. Relevant equations



    3. The attempt at a solution

    How do I even know that w ε S? That's a huge supposition isn't it? Well to prove that one set is a subset of the other set, all you need to do is prove that each share a common element right?

    And we know that the Set S has the following elements:

    1. v+w [from the last proof]
    2. v [as stated from our supposition that " v ε F^n is a particular solution to Ax=b"]
    3. x [as stated from our supposition " S subset F^n be the solution set to Ax=b"

    and S_0 contains the following elements in it's set:

    1. 0 [trivial solution]
    2. w [v+S_0) is defined as {v+w: w ε S_0}
    3. x [from our supposition "Let S_0 subset F^n be the solution set to the (homogenous) equation Ax=0."

    So since both have x in their elements, can't we say that S subset (v+S_0)?
     
    Last edited: Nov 22, 2011
  2. jcsd
  3. Nov 22, 2011 #2

    Mark44

    Staff: Mentor

    Unless S is empty, it has to have some elements. Any vector w in S satisfies Aw = b.
    No, not at all. Consider A = {1, 3, 5} and B = {1, 2, 3}. Both sets have 1 and 3 in common, but neither set is a subset of the other. For one set to be contained in another, each element of the first set must also be in the second set.
     
  4. Nov 22, 2011 #3


    Thank you but w is also in S_0. So it has to be the same value as S_0 right? How do we know if w ε S?
     
  5. Nov 22, 2011 #4

    Mark44

    Staff: Mentor

    Just so that other readers and I don't have to keep jumping from this page to the page you posted as a link, here is the problem statement. It would have been helpful for you to post it.
    1. Suppose F is a field, A [itex]\in[/itex] Mmn(F), b [itex]\in[/itex] Fm and v [itex]\in[/itex] Fn is a particular solution to the equation Ax = b. Let S0  [itex]\subseteq[/itex] Fn be the solution set to
    the (homogeneous) equation Ax = 0, and S [itex]\in[/itex] Fn be the solution set
    to Ax = b.
    (a) Prove that the set v + S0 = {v + w : w [itex]\in[/itex] S0} is a subset of S.
    (b) Prove that S is a subset of v + S0. (Hint: Suppose w [itex]\in[/itex] S. Show that w - v [itex]\in[/itex] S0. What does this say about w?)
     
  6. Nov 22, 2011 #5
    Sorry mate.
     
  7. Nov 22, 2011 #6

    Mark44

    Staff: Mentor

    The hint is "suppose w is in S0". Just go from there. All this means is that w is a vector such that Aw = 0.
     
  8. Nov 22, 2011 #7

    Mark44

    Staff: Mentor

    I'm getting the idea that you think you are providing a complete list of the elements of S. IOW, that S = {v+w, v, x}. That is not the case.

    In #1 above the hypothesis probably was something like, "suppose v and w are in S." Then you showed that v+w was also in S, thereby showing that S was closed under vector addition. v and w are symbols that can represent any of a possibly infinite number of vectors.
     
  9. Nov 22, 2011 #8
    Here's what i know now:

    A ε M_mn (F) where F is a field
    b ε F^m
    v ε F^n [is a solution to Ax=b]
    x ε F^n [is a solution to both Ax=b and Ax=0]
    w ε F^n [since the definition states v+S_0 = {v+w: w ε S_0}]

    S_0 is the solution set to equation Ax=0
    S is the solution to Ax=b

    I also know that A(v+w)=b. Then from there it follows that (v+w) is an element of S

    But i'm not sure if i can say w ε S since (v+w) is an element of S.

    I'm not able to make that next step to come to that conclusion. This is where i'm stuck at.


    Is it because (v+w) ε S, and S is a subset of F^n, then from the field axiom v ε F and w ε F?
     
  10. Nov 22, 2011 #9

    Mark44

    Staff: Mentor

    Of course you can assume that w ∈ S. That's what "Suppose w ∈ S" means.
     
    Last edited: Nov 22, 2011
  11. Nov 23, 2011 #10

    vela

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    Fine up to here.
    No, what you wrote would imply that b=0.

    The problem statement is saying that S is the solution set to the equation Ax=b. In other words, [itex]S = \{x \in F^n : Ax=b\}[/itex]. Similarly, you have [itex]S_0 = \{x \in F^n : Ax=0\}[/itex].
    This isn't really a definition of w, which is really just a dummy variable here. The problem statement is saying the notation "v+S0" means "the set of vectors of the form v+w where w is some vector in S0." You could have just as well written [itex]v+S_0 = \{v+Z : Z \in S_0\}[/itex] That no more defines Z as the original definition defines w. Both w and Z are dummy variables. They have no meaning outside of the context of the definition of v+S0.
    You have to somehow define w first. Note that the w in part (a) and the w in part (b) have absolutely nothing to do with each other.
    You seem to be unclear on what you can assume.

    To show S is a subset of v+S0, what you need to show is that for all vectors w in Fn, if w is in S, it is also in v+S0. So you assume w is in S and then show it follows that w is an element of v+S0. (If w is not in S, it doesn't matter because the implication is then automatically true.)
     
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