$\dim (V_1+V_2) \geq \dim(V_1 \cap V_2) +2$ I am trying to solve this problems from an old qual exam. I know a way to 
prove this but I feel like its too long for something that looks pretty simple.
Can anybody suggest a cleaner way? for example that uses the dimension formula
for subspaces $ \dim (V_1) + \dim(V_2)=\dim(V_1+V_2)+ \dim(V_1\cap V_2)$
Here is the statement.
Let $V$ be a finite dimensional vector space over a field $F$. Let $V_1$ and $V_2$
be subspaces of $V$ such that $V_1 \nsubseteq V_2$ and $V_2 \nsubseteq V_1$. Prove that $\dim (V_1+V_2) \geq \dim(V_1 \cap V_2) +2$
My solution involves the following facts: $V_1 \cap V_2 \subseteq V_i \subseteq V_1+V_2$ and then by way of contradiction assume the statement is false and proceeding. But this took me about a page to arrive at the desired contradictions
using various cases etc. In particular I did not explicitly use the dimension formula. I feel this is way too much for a problem that looks like a simple application of the dimension formula (?). Can anybody suggest a quicker way?
Thanks for all your help.
 A: Since $\dim(V_{1}),\dim(V_{2}) > \dim(V_{1}\cap V_{2})$, the dimension formula gives you the desired result.
To see the first claim I make, observe that since none of the subspaces contain the other, $V_{1}$ is strictly larger than the intersection and so is $V_{2}$
In gory details,
$$\dim(V_{1}) \geq \dim(V_{1}\cap V_{2}) + 1$$
$$\dim(V_{2}) \geq \dim(V_{1}\cap V_{2}) + 1$$
Thus, $\dim(V_{1}) + \dim(V_{2}) \geq 2\dim(V_{1}\cap V_{2}) + 2$.
Now use the dimension formula.
A: One has $$\dim(V_1+V_2)>\max(\dim(V_1),\dim(V_2))\geq\min(\dim(V_1),\dim(V_2))>\dim(V_1\cap V_2),$$
where the outer inequalities are strict because of the hypotheses $V_1\not\subseteq V_2\not\subseteq V_1$, which imply $V_1+V_2\supsetneq V_i$ and $V_i\supsetneq V_1\cap V_2$ for $i=1,2$) (one also uses the fact that everything is finite dimensional). With two of the inequalities strict, the outer terms differ by at least$~2$; you can easily see how to get a case where they differ by exaclty$~2$.
A: Take a basis for $V_1 \cap V_2$. Since $V_1 \subsetneq V_2$ and $V_2 \subsetneq V_1$, we have a vector $v$ in $V_1$ and not in $V_2$, and a vector $w$ in $V_1$ and not in $V_2$. We can expand our basis for $V_1 \cap V_2$ by adding $v$ and $w$. You should be able to show that this is still linearly independent. What does this imply about the dimension of $V_1 + V_2$ which contains this linearly independent set?
