Sum of vector subspaces I am trying to self study linear algebra and am stuck on a problem. It comes from Axler's Linear Algebra done right example 1.38 and I don't understand the solution that I could find online. 

Suppose that $U=\{(x,x,y,y)\in F^4 : x,y \in F\}$ and $W=\{(x,x,x,y) \in F^4 : x,y \in F\}$, show that $U+W=\{(x,x,y,z) \in F^4 : x,y,z \in F\}$.

Its stated that the sum of vector subspaces is the set of all possible sums of the elements but how does $x+y=y$ and $y+y=z$
 A: There's a slight misunderstanding with how you've read the question, arguably stemming from how Axler has labelled his variables. Here's a more verbose but perhaps clearer version of the question:
Let $x, y, z \in F$. Show that there exists $a, b, c, d \in F$ such that $(x, x, y, z) = (a, a, b, b) + (c, c, c, d)$. Similarly, let $a, b, c, d \in F$ and show that there exists $x, y, z \in F$ such that $(a, a, b, b) + (c, c, c, d) = (x, x, y, z)$.
A: You are getting hung up on the notation. The variables $x$, $y$, and $z$ used to define the sets $U$, $W$, and $\{(x,x,y,z)\in F^4 : x,y,z\in F\}$ have nothing to do with one another.
We must show that $U+W = S$, where $S:=\{(x,x,y,z)\in F^4 : x,y,z\in F\}$.
Let's first show "$\subseteq$": Fix $u:=(x,x,y,y)$ in $U$ and $w:=(a,a,a,b)$ in $W$. Then
$$
u+w = (x+a,x+a,y+a,y+b)
$$
is in $S$ because $x+a=x+a$, $y+a$, and $y+b$ are all elements of $F$.
Now we prove "$\supseteq$": Fix $s=(x,x,y,z)$ in $S$. We need to find $u\in U$ and $w\in W$ such that $s=u+w$. Consider $u:=(x-y,x-y,0,0)$ and $w:=(y,y,y,z)$. Then $u\in U$, $w\in W$, and
$$
s=(x,x,y,z)=(x-y,x-y,0,0)+(y,y,y,z)=u+w\in U+W.
$$
Therefore $U+W=\{(x,x,y,z)\in F^4 : x,y,z\in F\}$.
