Finding a basis for two subspaces of $\mathbb {R}^4$ Let $ U= \{(x_1,x_2,x_3,x_4)\mid x_2+x_3=0 , x_1-4x_4=0\}$ , similarly $V=\{(x_1,x_2,x_3,x_4)\mid x_1=0\}$.
How can I find a basis for each of the subspaces and how can one determine the dimension of their sum/intersection ?
I just started studying linear algebra and I'm really having a tough time trying to find a basis for a vector space.
 A: You can think about dimension as the minimum number of free parameters with which you can determine every vector of a space. So, for $U$, you know that $x_3=-x_2$ and $x_1 = 4x_4$. Then, 
$$
U=\{(4x_4,x_2,-x_2,x_4)\,|\,x_2,x_4\in\mathbb{R}\} = \operatorname{Span}\{(4,0,0,1),(0,1,-1,0)\}.
$$
Analogously, for $V$:
$$
V= \{(0,x_2,x_3,x_4)\,|\, x_2,x_3,x_4\in\mathbb{R}\} = \operatorname{Span}\{(0,1,0,0),(0,0,1,0),(0,0,0,1)\}.
$$
If a vector is in the intersection, then its first component must be zero, because it's in $V$. On the other hand, if it's in $U$ its fourth component must also be zero and $x_2=-x_3$. Then:
$$
U\cap V = \operatorname{Span}\{(0,1,-1,0)\}.
$$
A: Space $U$ is comprised of all vectors of the form
$$(x_1,x_2,-x_2,\frac14x_1)$$
thus every vectorin $U$  is generated by two basis vectors:
$$(x_1,x_2,-x_2,\frac14x_1) = x_1(1,0,0,\frac14)+x_2(0,1,-1,0)$$
This means that 
$e_1 = (1,0,0,\frac14)$ and $e_2 = (0,1,-1,0)$ and dimension of $U$, $dimU=2$.
Similarly the $dimV=3$.

Intersection
$I = U \cap V = (0,x_2,-x_2,0)$, $dimI=1$, we need only one vector to generate $I$.
Sum
$S = U + V$= we need three vectors to generate all vectors of the form $(x_1,x_2,x_3,0)$, but also we need sometimes $x_4$ thus 4th vector. $dimS=4$, we need 4 independent vectors to generate it.
