1. Subspaces and Dimension

Let V be vector space with dim(V)=n. And let U,W be subspaces with n-1 dimension each,I need to prove that: U+W=V if and only if U!=W

Thank you all!

2. I'll assume you know how to prove that if $U+W=V$ then $U\neq W$ (it is trivial).

Assume U,W are subspaces of V with $dim V = n, ~ dimU = dim W = n-1$ such that $U \neq W$.

Let $B = \{w_1,w_2,...,w_{n-1}\}$ be a basis for W.

Let $u \in U - W ~ \text{s.t.} ~ u \neq 0$ (we know one such element exists otherwise $U=W$ since $0\in U, ~ 0 \in W$).
Then u is not spanned by $\{w_1,w_2,...,w_{n-1}\}$, otherwise it would be an element of W (by definition). Then, $B' = \{w_1,w_2,...,w_{n-1},u\}$ is an independent set and obviously $dim B' = n$ and therefore it is a basis of V.

This gives us that $V \subset U+W$.
However, obviously there can be no element in $U+W$ that is not in V since they are subspaces, therefore $U+W \subset V \rightarrow U+W = V$

By the way, what uni are you studying in?

3. Thanks a lot! But, your assumption is wrong, can you prove the first line please?

4. Well, assume $U+W=V$. Assume, by contradiction, that $U = W$. Then $U+W = U = W = V$

What does this give us? (hint: look at the dimensions)

5. That the dimension of U and W is not the dimension of V. Paradox?
6. Yes, that is correct! since $U+W=V$, we must have that $dim(U+W) = dim(V)$ but in this case $dim(U+W)=n-1$, $dim(V)=n$ therefore we reach a contradiction, and then $U \neq W$.