"Linear Algebra Done Right" Sum of Subspaces Question I'm currently reading Axler's "Linear Algebra Done Right" and I have encountered the following example, which I have trouble verifying:
Suppose that $U = \{(x,x,y,y)\in \mathbf{F}^4: x,y \in \mathbf{F}\}$ and $W = \{(x,x,x,y)\in \mathbf{F}^4: x,y \in \mathbf{F}\}$. Then, $U+W = \{(x,x,y,z)\in \mathbf{F}^4 : x,y,z \in \mathbf{F}\}$
I don't understand how when you add the 3rd coordinates $x+y$ it's $y$. Are these y's different? Like when you add 3rd coordinates $2+1$ is the same as $3$. Because in that case, $U+W = \{(x,x,x,z)\in \mathbf{F}^4 : x,y,z \in \mathbf{F}\}$ wouldn't be also valid? Also, where did the $z$ come from? Does this stem from the same notion as said before?
I'm still new to the subject, so a clear intuitive explanation would be very much appreciated. Thanks!
 A: Let's be a little more careful in reading the notation.  First,
$$ \newcommand{\F}{\mathbf{F}} U := \{ (x,x,y,y) \in \F^4 : x,y\in\F \}. $$
This means that $U$ consists of all of the vectors in $\F^4$ where the first two coordinates are the same and and the last two coordinates are the same.  The variables are $x$ and $y$ are placeholders.  Similarly,
$$ V := \{ (x,x,x,y) \in \F^4 : x,y\in \F \} $$
is the set of all vectors where the first three coordinates are the same.  Again, the $x$ and $y$ are placeholder variables.  To ease the confusion, let's use another pair of letters instead, say $\xi$ and $\eta$.  We could then write
$$ V := \{ (\xi, \xi, \xi, \eta) \in \F^4 : \xi, \eta \in \F^4 \}. $$
By definition, the sum of these two spaces consists of all linear combinations of elements from each.  That is,
$$ U+V := \{ au + bv : a,b\in\F, u\in U, v\in V\}. $$
So suppose that $z = au + bv$ is a typical element of $U+V$ with $u = (x,x,y,y)$ and $v = (\xi, \xi, \xi, \eta)$.  Then
\begin{align}
z
&= au + bv \\
&= (ax,ax,ay,ay) + (b\xi, b\xi, b\xi,b\eta) \\
&= (\underbrace{ax+b\xi}_{\alpha}, \underbrace{ax+b\xi}_{\alpha}, \underbrace{ay+b\xi}_{\beta}, \underbrace{ay+b\eta}_{\gamma}) \\
&= (\alpha, \alpha, \beta, \gamma). \end{align}
In other words, if $z \in U+V$, then the first two coordinates of $z$ must agree, but the second two may not agree with either each other or the first two coordinates.  Moreover, since $a$, $b$, $x$, $y$, $\xi$, and $\eta$ are free to vary over $\F$, it follows that $\alpha$, $\beta$, and $\gamma$ are similarly free to vary over over $\F$.  Hence we could write
$$ U+V = \{ (\alpha, \alpha, \beta, \gamma) \in \F^4 : \alpha, \beta, \gamma\in \F \}. $$
Notice, however, that this is exactly the same as saying that
$$ U+V = \{ (x,x,y,z) \in \F^4 : x,y,z \in \F\}. $$
However, if we use $x,y,z$ as placeholder variables everywhere, then we don't have to introduce a bunch of extra letters.  Since the mathematical context is unambiguous and since we have few enough symbols already, this kind of recycling is common (and, indeed, is good practice).
A: The notation means, for instance, that
$$
(1,1,0,0)\in U, \qquad (1,1,1,0)\notin U
$$
More generally, a vector $(a,b,c,d)$ belongs to $U$ if and only if $a=b$ and $c=d$. Similarly,
$$
(1,1,1,0)\in W,\qquad (1,1,0,0)\notin W
$$
More generally, a vector $(a,b,c,d)$ belongs to $W$ if and only if $a=b=c$.
Suppose a vector $(a,b,c,d)$ belongs to $U$; thus $a=b$ and $c=d$, so
$$
(a,b,c,d)=(a,a,c,c)=a(1,1,0,0)+c(0,0,1,1)
$$
Since, clearly, $(1,1,0,0),(0,0,1,1)\in U$, we can say that
$$
U=\operatorname{span}\{(1,1,0,0),(0,0,1,1)\}
$$
Similarly,
$$
W=\operatorname{span}\{(1,1,1,0),(0,0,0,1)\}
$$
Note that the two vector sets we found are also linearly independent.
Hence
$$
U+W=\operatorname{span}\{(1,1,0,0),(0,0,1,1),(1,1,1,0),(0,0,0,1)\}
$$
How can we find a basis for $U+W$? A good method is doing Gaussian elimination on the matrix
$$
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
$$
The elimination goes as follows
\begin{align}
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
&\xrightarrow{R_3\gets R_3-R_1}
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}
\xrightarrow{R_3\gets R_3-R_2}
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & -1 \\
0 & 0 & 0 & 1
\end{bmatrix}
\\[6px]
&\xrightarrow{\substack{R_4\gets R_4+R_1\\R_3\gets -R_3}}
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\xrightarrow{R_2\gets R_2-R_3}
\begin{bmatrix}
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0
\end{bmatrix}
\end{align}
Row operations don't change the row space, so we can deduce that
$$
U+W=\operatorname{span}\{(1,1,0,0),(0,0,1,0),(0,0,0,1)\}
$$
(and the three vectors form a linearly independent set). Therefore the vectors in $U+W$ are (uniquely) linear combinations of the form
$$
x(1,1,0,0)+y(0,0,1,0)+z(0,0,0,1)=(x,x,y,z)
$$
as $x,y,z\in\mathbf{F}$; in a different, but equivalent notation,
$$
U+W=\{(x,x,y,z):x,y,z\in\mathbb{F}\}
$$
The actual name of the letters is irrelevant: the set above is the same as
$$
\{(a,a,b,c):a,b,c\in\mathbb{F}\}
$$
This just means that a vector belongs to $U+W$ if and only if its first two components are equal.
