I have to find a subspace $W$ of $F^4$ such that $F^4=U \oplus W$.
$U:=\{(x,x,y,y) \in F^4:x,y \in F\}$
I found the topic with the same problem. The solution was:
$W:=\{(0,b-a,c-d,0)\in F^4:a,b,c,d \in F\}$, because every vector of $F^4$ we can write as:
$(a,b,c,d)=(a,b-a+a,c-d+d,d)=(a,a,d,d)+(0,b-a,c-d,0)$.
I used the similar strategy, but my $W$ is another one. So my solutions is:
$(a,b,c,d)=(a,a,b,b)+(0,b-a,c-b,d-b)$. Therefore I have two questions.
Is my $W$ correct?
If yes, then subspaces $U$ and $W$ must not be unique? So I can have infinitely many such $W$, which will work for direct sum? Or maybe not infinitely many, but just many?
Actually, you did not state clearly which space your $W$ is, but I will assume that it is$$W=\left\{(0,b−a,c−b,d−b)\mid a,b,c,d\in F\right\}.$$No, this is not right. Take, say, $a=b=0$ and $c=d=1$. Then$$\overbrace{(0,b−a,c−b,d−b)}^{\phantom W\in W}=(0,0,1,1)\in U\setminus\{(0,0,0,0)\},$$and therefore $U\cap W\ne(0,0,0,0)$.
But, yes, there are infinitely many solutions.