TITLE: Dimension of the sum of two vector subspaces
QUESTION [18 upvotes]: $\dim(U_1+U_2) = \dim U_1 +\dim U_2 - \dim(U_1\cap U_2).$
I want to make sure that my intuition is correct. Suppose we have two planes $U_1,U_2$ though the origin in $\mathbb{R^3}$. Since the planes meet at the origin, they also intersect, which in this case is a one-dimensional line in $\mathbb{R^3}$. To obtain the dimension of $U_1$ and $U_2$, we add the dimensions of the planes (4), and the subtract the dimensions of the line (1), which results in (3). 
*additional question(s): 
Can we generalize this notion to $\mathbb{F^{n}}$?
Suppose we have an additional case where $U_1$ and $U_2$ are planes in $\mathbb{R^3}$, but $U_1 \subseteq U_2$. In this instance, $dim(U_1 + U_2) < 3$, because the first two-dimensional plane is contained in the second and as a result, the dimensions of the subspaces when summed cannot exceed two. Since both subspaces $U_1,U_2$ are two dimensional and $U_1 \subseteq U_2$, then their intersection is also two-dimensional, concluding $dim(U_1+U_2)=2+2-2 = 2$.
Is this proper intuition?

REPLY [13 votes]: Cameron Buie's answer does answer the original question sufficiently. However, I'm adding the formal proof of the theorem in context (taken from "Linear Algebra Done Right" by Sheldon Axler).

If $U_1$ and $U_2$ are subspaces of a finite dimensional vector space then: $$\dim(U_1+U_2)=\dim U_1 + \dim U_2 - \dim(U_1 \cap
 U_2)$$

Let $u_1,...,u_m$ be a basis of $U_1\cap U_2$; thus $\dim (U_1\cap U_2)=m$. Because $u_1,...,u_m$ is a basis in $U_1\cap U_2$, it is linearly independent in $U_1$. Hence this list can be extended to a basis $u_1,...,u_m,v_1,...,v_j$ of $U_1$. Thus, $\dim U_1 = m+j$. Also extend $u_1,..,u_m$ to a basis $u_1,...,u_m, w_1,...,w_k$ of $U_2$. $\dim U_2 = m +k$.
We will show that $u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$ is a basis of $U_1+U_2$. This will complete the proof because then we will have $$\dim(U_1+U_2)=m+j+k=\dim U_1 + \dim U_2 - \dim (U_1\cap U_2)$$
We just need to show that the list $u_1,...,u_m,v_1,...,v_j,w_1,...,w_k$ is linearly independent. To prove this, suppose: 
$$a_1u_1+...+au_m+b_1v_1+...+b_jv_j+c_1w_1+...+c_kw_k=0$$
where all $a,b,c$'s are scalars. We need to show that all the $a,b$ and $c$'s are $0$.
The equation can be rewritten as 
$$c_1w_1+...+c_kw_k=-a_1u_1 - ... -a_mu_m -b_1v_1 - ... - b_j v_j$$
Which shows that $c_1w_1+...+c_kw_k\in U_1$. But actually all $w$'s are in $U_2$. So the LHS must be an element of $U_1\cap U_2$.
$c_1w_1+...+c_kw_k=d_1u_1+...+d_mu_m$ for some choice of scalars $d_1,d_2,...,d_m$. But $u_1,...,u_m,w_1,...,w_k$ is linearly independent. So our last equation implies that all the $c$'s equal $0$.
Thus our original equation involving $a,b,c$ becomes
$$a_1u_1+...+a_mu_m+b_1v_1+...+b_jv_j=0$$
But we already knew that the list $u_1,...,u_m,v_1,...,v_j$ is linearly independent. This equation implies that all the $a$'s and $b$'s are $0$. We now know that all $a,b$ and $c$'s are $0$, hence proving our original claim.