Prove $\dim W \ge 2$ 
Let $U_1, U_2, W$ subspaces of a finite dimensional vector space, such that:  
  
  
*
  
*$U_1 \cap U_2 = \{0\}$
  
*$U_1 \cap W \ne \{0\}$
  
*$U_2 \cap W \ne \{0\}$
  
  
  Show that $\dim W \ge 2$.

My proof: 
There's $v_1 \in U_1 \cap W$ and $v_2 \in U_2 \cap W$.  
If we show that $\{v_1, v_2\}$ is linearly independent set, then the dimension of $W$ must be at least $2$.
Indeed, Assume that $\alpha_1 v_1 + \alpha_2 v_2 = 0 \Rightarrow \alpha_1 v_1 = -\alpha_2v_2$
So, we have $\alpha_1v_1 = v = \alpha_2v_2$ which is a contradiction to $U_1 \cap U_2 = \{0\}$.
And so, $\dim W \ge 2$.
I'd be glad to get a review of my work. Is there a more direct approach? Am I being rigorous? 
Thanks.  
 A: Your proof is almost fine:


*

*You need to assume $v_1\ne 0$ and $v_2\ne 0$.

*There is no contradiction. You get $v \in U_1 \cap U_2 = \{0\}$ and so $\alpha_1=0=\alpha_2$, which is what you need to prove that $v_1$ and $v_2$ are linearly independent.
A: 
"Assume that $\alpha_1 v_1 + \alpha_2 v_2 = 0 \Rightarrow \alpha_1 v_1 = -\alpha_2v_2$".

What you want to write is "Assume $\alpha_1 v_1 + \alpha_2 v_2 = 0$, then $\alpha_1 v_1 = -\alpha_2v_2$". It's very different, because you can't parse a mathematical symbol like $\implies$ as if it were natural language. What you wrote means 'assume the conditional statement holds' and that's not what you want. 
Another thing that should be made clearer is that you should take $\alpha _1, \alpha_2$ arbitrarily in the ground field if you're planning to prove that $\forall \alpha _1, \alpha_2(\alpha_1v_1+\alpha _2v_2=0\implies \alpha _1=0=\alpha _2)$. If, on the other hand, you want to assume that $\{v_1, v_2\}$ is a linearly dependent set and get a contradiction, then you should write something like "Assume there exist $\alpha _1, \alpha_2$ such that $\alpha_1v_1+\alpha _2v_2=0$ and $\alpha _1\neq 0\lor \alpha_2\neq 0$". Since you didn't quantify over $\alpha_1, \alpha_2$ at all, it's not clear what you're doing.

"So, we have $\alpha_1v_1 = v = \alpha_2v_2$"

No, correct would be "So, we have $\alpha_1v_1 = \color{red}-\alpha_2v_2$". If you want to include that '$v$', you need to say what it is, so you can write something like "So, we have $\alpha_1v_1 = v = -\alpha_2v_2$, for some $v\in \left(U_1\cap W\right)\cap \left(U_2\cap W\right)=U_1\cap U_2\cap W=\{0\}$". Now you want to conclude that $\alpha _1=\alpha _2=0$, but you can't without the assumption that $v_1\neq 0\neq v_2$ which you should make at the beginning of the proof. Then you can conclude without the contradiction.

Alternatively assume $\dim \left(W\right)\in \{0,1\}$.
If $\dim \left(W\right)=0$, then $U_1\cap W=\{0\}$.
If $\dim \left(W\right)=1$, then since $\{0\}\subset U_1\cap W\subseteq W$, necessarily $0< \dim\left(U_1\cap W\right)\leq 1$ and thus $U_1\cap W=W$. Similarly $U_2\cap W=W$. From this it follows that $W=U_1\cap U_2\cap W\subseteq U_1\cap U_2$ and consequently $U_1\cap U_2\neq \{0\}$.
