1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Intersections of Subspaces and Addition of Subspaces

  1. Jan 22, 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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Can you offer guidance or do you also need help?



Similar Discussions: Intersections of Subspaces and Addition of Subspaces
  1. Sums of subspaces? (Replies: 0)

Loading...