Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

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.

share|cite|improve this question
Find two things in it. Unless you are very unlucky, their sum won't be in it. Or, find one thing in it. Unless you are very unlucky, the negative of that thing won't be in it. – Gerry Myerson Sep 21 '12 at 5:58
A question is not a statement; to "prove a question" makes no sense. – joriki Sep 21 '12 at 6:00
@GerryMyerson: The quickest thing to find is $(0,0,0)$, which is going to make him rather unlucky either way. – Michael Joyce Sep 21 '12 at 6:53

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 $(x_1+y_1)^2=x_2+y_2$?

share|cite|improve this answer

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.

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.