# Intersection and Addition of Subspaces

1. Jan 23, 2012

### TranscendArcu

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 $S = \left\{ S_1,...,S_n \right\}$. 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.

Last edited by a moderator: May 5, 2017
2. Jan 23, 2012

### 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.

3. Jan 23, 2012

### HallsofIvy

Staff Emeritus
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.