Find three nontrivial subspaces such that $\mathbf{F}^5=U\oplus W_1\oplus W_2\oplus W_3$, where $U=\{(x, y, x+y, x-y, 2x): x, y\in \mathbf{F}\}$. The Problem: Find three subspaces  $W_1, W_2, W_3$ of $\mathbf{F}^5$, none of which equals $\{0\}$, such that $\mathbf{F}^5=U\oplus W_1\oplus W_2\oplus W_3$, where $U=\{(x, y, x+y, x-y, 2x)\in\mathbf{F}^5: x, y\in \mathbf{F}\}$. $\mathbf{F}$ is a field.
Source: Sheldon Axler's Linear Algebra Done right, thrid edition.
My Background: Selected topics from Chapter 1-4 of Stephen H. Friedberg, Arnold J. lnsel, Lawrence E. Spence's Linear Algebra, 5th edition from a year ago.
My Attempt: The problem seems very straightforward, but I am somehow dead stuck. Clearly we want to find a general expression for the vectors $\vec{w_i}$ in each $W_i$ (like $(x, y, x+y, x-y, 2x)$ for $U$) so that each $\vec{x}\in\mathbf{F}^5$ can be expressed uniquely as $\vec{x}=\vec{y}+\vec{w_1}+\vec{w_2}+\vec{w_3}$, where $\vec{y}\in U, \vec{w_1}\in W_1$, etc. But just where exactly do we begin? Indeed we can let $(a, b, c, d, e)\in\mathbf{F}^5$ be arbitrary, but then how do we establish a series of equations to determine the general expression for the elements in each $W_i$?
Any help would be greatly appreciated.
 A: A basis for $U$ is $\{(1,0,1,1,2),(0,1,1,-1,0)\}$. To find three non trivial subspaces we just need to choose three different vectors $v_i\not\in U$ such as $span(v_i)\not = span(v_j)$ when $i\not =j$ For example $v_i=e_i$ for $i=1,2,3$
A: Let's start with a short analysis of the problem
$\mathbb F^5$ is a vector space of dimension equal to 5 over $\mathbb F$. Denote by $C = \{e_1, \dots, e_5\}$ the canonical basis. The two vectors $u_1 = (1, 0 ,1, 1, 2)^t$ and $u_2 = (0, 1, 1,-1, 0)^t$ are linearly independent and span $U$.
Therefore $W_1, W_2,W_2$ have all to be of dimension equal to one.
Incomplete basis theorem
A theorem states that if $L$ is a family of linearly independent vectors, and $G$ is a family of vectors that spans the vector space $V$, then you can pick up a subset $F \subseteq G \setminus L$ such that $L \cup F$ is a basis of $V$.
Let's apply this to our problem with $L = \{u_1, u_2\}$ and $G=C$. We have to find a subset $\overline{C}$ of three vectors of $C$ such that $L \cup \overline{C}$ is a basis of $\mathbb F^5$.
And we can see that it is the case of $\{e_3, e_4, e_5\}$ as the determinant of $\lvert u_1, u_2 , e_3, e_4, e_5 \rvert$ is equal to one and therefore not zero.
Finally
Just take $W_1 = \mathbb F e_3$, $W_2 = \mathbb F e_4$ and $W_3 = \mathbb F e_5$.
