Addition of 2 subspaces. The questions goes like this:
Write down the subspace $U + V$ explicitly if
$U = {(t, 2t, 3t)}$ and $V = {(t, -2t, 0)}$ where $t$ is a member of the real number
My answer for the question is:
$U + V$ $=$ {$(x + y, 2x - 2y, 3x)$ | $x$, $y$ are member of the real number}
However, I am confused as other have mentioned that the answer should be instead:
$U + V\{(2t, 0, 3t)\mid t \text{ is a member of the real number}\}$
In my opinion, the I feel that the second answer is wrong as the two sub-spaces are not a subset of each other. Is my thinking wrong?
 A: The first answer is the correct one. The reason being that $t$ is a dummy variable used to represent both linear subspaces $U, V$, which are lines by the way.
When you make the sum, there is no link between those dummy variables that have to be separated. So using two different variables $x,y$ as in the first answer is the correct approach.
A: When we say $l$ is in the set $S=\{f(x)\mid x\in A\}$, we mean by $l$ takes the form $f(x)$. More precisely, $l=f(x)$ for some $x\in A$.
Next we define the set $U\oplus V$ by $$x\in U\oplus V \Longleftrightarrow x=y+z \mbox{ for some } y\in U,z\in V.$$ But then the $y,z$ from $U,V$ have their own form, namely $(t, 2t, 3t)$ and $(t, -2t, 0)$ for some real number $t$. (To not confuse ourselves we use different dummy variable, say $r,s$.) $$\Longleftrightarrow x=(r, 2r, 3r)+(s, -2s, 0) \mbox{ for some }r,s\in\mathbb{R}$$
$$\Longleftrightarrow x=(r+s,2r-2s,3r) \mbox{ for some }r,s\in\mathbb{R}$$
So the set $U\oplus V$ is $\{(r+s,2r-2s,3r)\mid r,s\in\mathbb{R}\}$.
The point that you might have misunderstood is the duplicate misuse of dummy variables $r,s$.
Note that we may use $+$ instead of $\oplus$.
