Find a subspace $W$ of $\mathbb{F}^4$ such that... - checking my answer! The question is the same as one previously asked, but I can't comment so I had to ask my own. Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U \oplus W$
Suppose $U=\{ (u_1,u_1,u_2,u_2)\in \mathbb{F}^4:u_1,u_2\in \mathbb{F}\}$.
Find a subspace $W$ of $\mathbb{F}^4$ such that $\mathbb{F}^4 = U\oplus W$.
I chose $W = \{ (0,w_1,0,w_2)\in \mathbb{F}^4:w_1,w_2\in \mathbb{F}\}$.
Is this incorrect or can I use it just as previous answers have been $(0,w_1,w_2,0)$?
From what I can tell, my answer allows $U\cap W = \{0\}$ and $U+W=\{(u_1,u_1+w_1,w_1,u_2+w_2)\in\mathbb{F}^4:u_1,u_2,w_1,w_2\in\mathbb{F}\}=\mathbb{F}^4$.  
Is this a sufficient argument?

The example you provide doesn't work. The vector $(1,1,1,1)\in \mathbb{F}^4$ but can't be written as $(x,2x,y,2y)$. But you can still solve your problem by taking the orthogonal complement $W^{\bot}$ of $W$. With $W$ as defined, its orthogonal complement is given by:$$W^{\bot}=\left\lbrace X=(x,y,z,t)\in\mathbb{F}^4\mid\,\langle X,v\rangle=0,\mbox{ for all }v\in W \right\rbrace ,$$
where $\langle \cdot,\cdot\rangle$ represents the usual inner product of $\mathbb{F}^4$.
On the orther hand, we can write 
$$W=\mbox{span}
\left\lbrace
\begin{pmatrix}
1\\
1\\
0\\
0
\end{pmatrix}
,
\begin{pmatrix}
0\\
0\\
1\\
1
\end{pmatrix}
\right\rbrace
.$$ 
Therefore $X=(x,y,z,t)\in W^{\bot}$ if and only if:
\begin{align*}
\left\{
\begin{array}{l}
\overline{x}=-\overline{y}\\
\overline{z}=-\overline{t}
\end{array}
\right.
\end{align*}
This system provides
$$W^{\bot}=\mbox{span}
\left\lbrace
\begin{pmatrix}
-1\\
1\\
0\\
0
\end{pmatrix}
,
\begin{pmatrix}
0\\
0\\
-1\\
1
\end{pmatrix}
\right\rbrace
.$$ 
You can immediately check that $W\oplus W^{\bot}=\mathbb{F}^4$.
