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The definition of sum of subspaces in Axler Sheldon book "Linear Algebra Done Right" is:

The sum of $U_1,\dots,U_m,$ denoted $U_1+\cdots+U_m$, is defined to be the set of all possible sums of elements of $U_1,\dots,U_m$. More precisely, $$U_1 + \cdots + U_m = \{u_1 + \cdots + u_m : u_1 \in U_i,\dots, u_m \in U_m\}.$$

Right after the definition, the following example is provided:

Suppose $U$ and $W$ are subspaces of $F^3$ given by $$ U = \{(x,0,0) \in F^3 : x \in F\} \text{ and } W = \{(0,y,0)\in F^3: y \in F\}.$$ then \begin{equation}\tag{1}\label{1}U+W = \{(x,y,0) : x,y \in F\}.\end{equation} Then let $V$ be $$V = \{(y,y,0)\in F^3: y \in F\}.$$ the book says that $U+V$ is still given by equation \eqref{1}. I am confused with the previous statement, as I would expect that $U+V = \{(x+y,y,0):x,y\in F\}.$ I would appreciate an explanation of this example.

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    $\begingroup$ They are the same sets, the representation is slightly different. Choose $x'=x+y, y'=y$ then $x=x'-y', y=y'$. $\endgroup$ – copper.hat Sep 26 '17 at 19:08
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You are right that if you directly apply the definition, then $U+V = \{(x+y,y,0):x,y\in F\}$. However, $\{(x,y,0):x,y\in F\}$ and $\{(x+y,y,0):x,y\in F\}$ are the same set! Given any $(x,y,0)$ in the first set, if we let $z=x-y$ then $(x,y,0)=(z+y,y,0)$ so it is also in the second set. And given any $(x+y,y,0)$ in the second set, if we let $z=x+y$ then $(x+y,y,0)=(z,y,0)$ so it is also in the first set.

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the basis of $U$ is $(1,0,0)$ and the basis of $V$ is $(1,1,0)$. and the basis of $U+V$ is $\{(1,0,0), (1,1,0)\}$ which is linearly equivalent to (the basis of $U+V$ can also be expressed as )$\{(1,0,0), (0,1,0)\}$.

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