Bases and dimensions of two spans Let $$U=\operatorname{span}\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$$ and $$ W=\operatorname{span}\{(1,2,2,-2),(2,3,2,-3),(1,3,4,-3)\}$$
I am supposed to find basis and dimensions for $U+W$ and $U\cap W$.  I know how to find a basis for a set of vectors (i.e., the basis for $U$ is $(1,1,0,-1),(1,2,3,0)$), but I do not know how to find $U+W$ and $U\cap W$.
Also, is the dimension for the basis of $U$ 2?  If I recall correctly, the dimension is equal to the number of vectors in the basis.
 A: Hint: You can easily prove that $U=\big\langle \underbrace{(1,1,0,-1)}_{u_1},\underbrace{(1,2,3,0)}_{u_2}\big\rangle$ and $W= \big \langle \underbrace{(1,2,2,-2)}_{w_1},\underbrace{(1,3,4,-3)}_{w_2}\big \rangle$.
If you want to find $U + W$, you have to check if all these above ($4$) vectors are linearly independent, which they are not. You can prove that only $3$ of them are linearly independent, which  can span $U+W$.
Now, if I'd want to find $U \cap W $, I would follow this safe way: 
First way: 
Let $\vec x \in U\cap W\implies \vec x \in U \text { AND }\vec x \in W$.
That means:
$\vec x= k\cdot ( 1,1,0,-1) + \lambda \cdot (1,2,3,0)$
$\vec x =\mu \cdot ( 1,2,2,-2) + \nu\cdot (1,3,-4,3)$, where $k,\lambda, \mu, \nu \in \mathbb R$.
So, you need to solve a $4\times 4 $ linear system with 4 variables: $k,\lambda, \mu,\nu \in \mathbb R$.
The only solution to the above system is $(k,\lambda,\mu,\nu)=(0,0,0,0)$.
That means $U\cap W= \{(0,0,0,0)\}$.
Second way: 
Since $u_1, u_2,w_1$ are linearly independent and $u_1,u_2,w_2$ are linearly independent as well, that means that no linear combination of $u_1, u_2$ can be written as linear combination of $w_1, w_2$ (but one specific case).
This means that there are no $k,\lambda,\mu, \nu \in \mathbb R$ and not all zero such that:
$k\cdot u_1 + \lambda \cdot u_2=\mu \cdot w_1 + \nu\cdot w_2$.
The only case this equation is true is when $k=\lambda=\mu=\nu=0$! 
Note: Practically, this means that $u_1,u_2,w_1,w_2$ are linearly independent!
