Intersection and Addition of Subspaces

[TranscendArcu]

Homework Statement

http://img824.imageshack.us/img824/3849/screenshot20120122at124.png

The Attempt at a Solution

Let S = \left\{ S_1,...,S_n \right\}. If L(S) = V, then T = \left\{ 0 \right\} and we are done because S + T = V. Suppose that L(S) ≠ V. Let B_1 \in T such that B_1 \notin L(S). Then the set Q =\left\{ S_1,...,S_n,B_1 \right\} is linearly independent. If L(Q) = V then we are done since S + T = V and S \cap T = \left\{ 0 \right\}. 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 Q' = \left\{ S_1,...,S_n,B_1,B_2 \right\} and B_1,B_2 \notin S. Then, if L(Q') = V, 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.

 

[kru_]

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.

 

[HallsofIvy]

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.