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$S = \{(x_1, x_2, x_3) \in \mathbb{R}^3\ |\ x_2 − (x_1)^2 = 0\}$

I found this question in an old exam and I'm not sure how to prove this question, but I know it is not a subspace. Any help is appreciated.

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    @GerryMyerson: The quickest thing to find is $(0,0,0)$, which is going to make him rather unlucky either way.2012-09-21

3 Answers 3

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You should start with checking whether it is closed under summation: if $(x_1,x_2,x_3),(y_1,y_2,y_3)\in S$, does it imply that $(x_1+y_1,x_2+y_2,x_3+y_3)\in S$. In this case, explicitly, this means that $x_1^2=x_2$ and $y_1^2=y_2$. Does this imply that $(x_1+y_1)^2=x_2+y_2$?

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General rule of thumb: if $S$ is of the form $S = \{x: Ax = 0\}$ then $A$ better be linear.

In the present case of $S = \{ (x_1,x_2,x_3) \in \mathbb{R}^3: x_2 - x_1^2 = 0\}$, consider $x = (1,1,0)$. Then $x \in S$ since $x_2 - x_1^2 = 1-1 = 0$. But what happens to $y = 2x$? We have that $y = 2x = (2,2,0)$ and thus $y_2 - y_1^2 = 2-4 = -2$ so that $y = 2x \notin S$. Hence $S$ cannot be a subspace because it is not closed under scalar multiplication.

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It is not a linear subspace, but it is an algebraic set.