Find an example of a nonempty subset $U$ of $\mathbb{R}^2$ where $U$ is closed under scalar multiplication but U is not a subspace of $\mathbb{R}^2$. I saw other questions asking the same and some of the answers were that any two lines through the origin work. But for my example I'm not sure if I'm doing it correct:
My example: Let $U = \left\{(x,y) : y = x \right\} \cup \left\{ (x, y) : y = -x\right\}$
To show this is not closed under vector addition let $a = (x_1 , x_2)$ and $b = (y_1, y_2)$ such that $x_2 = x_1$, $x_2 = -x_1$ and similar results for $y$.
Then $a + b = (x_1 + y_1, x_2 + y_2)$ such that $x_1 + y_1 = x_2 + y_2$ and $x_1 + y_1 = -x_2 – y_2$.
This seems closed under addition since all those values are $0$. Am I doing this completely wrong?
Best Answer
Your counterexample should be two specific vectors in $U$ (e.g., $(1,1),(1-1)$) instead of arbitrary vectors (i.e. $(x_1,x_2),(y_1,y_2)$).