Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$ Suppose $U = \{(x,x,y,y) \in \mathbb{F}^4 : x, y \in \mathbb{F}\}$ 
Find a subspace W of $\mathbb{F}^4$  such that $\mathbb{F}^4 =  U\oplus W$
Attempt:  Now from what I understand I would think that an element of $\mathbb{F}^4$ would look like $$(w,x,y,z) \quad\mbox{such that}\quad  
w,x,y,z \in \mathbb{F}$$
with that being the case I would use a subspace of the form: $$W = (w-x, 0, 0, z-y) \in \mathbb{F}^4 \quad\mbox{such that}\quad  w,x,y,z \in \mathbb{F}$$
But as a solution it was given that $$ W = (0,x,y,0). $$
Explanation?
I think I am not fully grasping how the direct sum sets are formed, but I got the idea that it was using an element from each subspace.
 A: Let $W=\{(0,w,z,0)\in\mathbb{F}^4:w,z\in\mathbb{F}\}$, then we check that such $W$ is the desired subspace, as the following two steps.


*

*Given $(a,b,c,d)\in\mathbb{F}^4$, it is easy to decompose the vector 
as below.
\begin{align}
(a,b,c,d)
&=(a,b-a+a,c-d+d,d)\\
&=(a,a,d,d)+(0,b-a,c-d,0),
\end{align}
where $(a,a,d,d)\in U$ and $(0,b-a,c-d,0)\in W$. Hence $\mathbb{F}^4=U+W$.

*If $(e,f,g,h)\in U\cap W$, then we have
$e=f$, $g=h$, $e=0$, and $h=0$. So
$$(e,f,g,h)=(0,0,0,0)$$
and hence $U\cap W=\{(0,0,0,0)\}$.

A: $ U $ is a subspace spanned by the linearly independent set $ S = \{ (1, 1, 0, 0), (0, 0, 1, 1) \} $. Therefore, it suffices to pick a subspace $ W $ which is spanned by two vectors such that their adjoinment to $ S $ would not disturb its linear independence. In other words, we need to extend $ S $ to a basis of $ \mathbb{F}^4 $.
It is easy to see that $ (1, 0, 0, 0), (0, 0, 0, 1) \notin U $. Now, we check if the set $ S' $ formed by adjoining these vectors to $ S $ is linearly independent. The standard method is to row reduce the matrix whose columns are elements of $ S $, but a more direct approach works here. Let $ s_i $ denote the elements of $ S' $:
$$ c_1 s_1 + c_2 s_2 + c_3 s_3 + c_4 s_4 = ( c_1 + c_3, c_1, c_2, c_2 + c_4) $$
For the left hand side to equal zero, it is then clear that we must have $ c_i = 0 $ for all coefficients, establishing linear independence of $ S' $. Therefore, $ S' $ is a basis of $ \mathbb{F}^4 $ (by the dimension theorem), and we may take $ W  = \textrm{span} \{(1, 0, 0, 0), (0, 0, 0, 1) \} = \{ (x, 0, 0, y) : x, y \in \mathbb{F} \} $.
