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Intersection and Addition of Subspaces

  1. Jan 23, 2012 #1
    1. The problem statement, all variables and given/known data

    http://img824.imageshack.us/img824/3849/screenshot20120122at124.png [Broken]

    3. The attempt at a solution
    Let [itex]S = \left\{ S_1,...,S_n \right\} [/itex]. If [itex]L(S) = V[/itex], then [itex]T = \left\{ 0 \right\}[/itex] and we are done because [itex]S + T = V[/itex]. Suppose that [itex]L(S) ≠ V[/itex]. Let [itex]B_1 \in T[/itex] such that [itex]B_1 \notin L(S)[/itex]. Then the set [itex]Q =\left\{ S_1,...,S_n,B_1 \right\}[/itex] is linearly independent. If [itex]L(Q) = V[/itex] then we are done since [itex]S + T = V[/itex] and [itex] S \cap T[/itex] [itex]= \left\{ 0 \right\} [/itex]. If not, then we may repeat the preceding argument beginning with the set Q. Thus, we would create a new set, call it Q', where [itex]Q' = \left\{ S_1,...,S_n,B_1,B_2 \right\}[/itex] and [itex]B_1,B_2 \notin S[/itex]. Then, if [itex]L(Q') = V[/itex], then we are done. If not, then we continue as above. This process must certainly end because V is itself stated to be finite. Thus, such a T as in the problem must exist.

    I think I'm going in the right direction, but I'm not sure if I'm executing the proof correctly.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Jan 23, 2012 #2
    I see where you are going, and you have the right concept. To be a bit clearer, how about starting with a basis for V, and let some part of that basis be the basis for S. Then go from there.
     
  4. Jan 23, 2012 #3

    HallsofIvy

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    Sorry, but TranscendArc's method is better. You can't simply assume that there exist a basis for V that includes a basis for S. It is true, of course, but TranscendArc's method proves that.
     
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