Let L be a vector space of dimension $ n $ and $ M,N \subset $ L linear subspaces such that $ dim(M)=n-1 $ and $ N \not \subset M $.
Prove that $ dim( M \, \cap N ) = dim(N) -1 $
My attempt:
Let $ (v_1,…,v_{n-1}) $ be a basis for $ M $ , we'll expand the basis of $ M $ to basis of $ M + N $ by the vectors $ (v_n,…,v_t) $ so that the basis of $ M + N $ is $ (v_1,…,v_t) $.
Since $ M + N \subseteq V $ then $ dim(M + N) \leq dim(V) $ , so that
$ n-1 \leq dim(M+N) = t \leq n $.
[ I dont know how to continue from here ]
Basically I want to use the dimension theorem for subspaces $ dim(M+N) = dim(M) + dim(N) – dim(M \cap N)$ But I got stuck and don't know how to continue, can you please help?
Edit:
As shown in the answers below, after showing that $ dim( M + N ) = n $ then, using the dimension theorem,
$ dim(M+N) = dim(M) + dim(N) – dim(M \cap N) \iff $ $ dim(M \cap N) = dim(M) + dim(N) – dim(M+N) \iff $ $ dim(M \cap N) = n-1 + dim(N) – n \iff dim(M \cap N) = dim(N) – 1 $ . And we've finished the proof.
Best Answer
If $\dim(M+N) = \dim(M)$, then since $M \subset M+N$, we have $M = M+N \supset N$ (contradicting $N \not\subset M$)
Therefore $\dim(M+N) > \dim(M) = n-1,$ which along with $\dim(M+N) \leq n$ implies $\dim(M+N) = n$.
Now finish by the dimension theorem.