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

I'm having trouble understanding the conditions of the set $$H = \{(x,y) \in \Bbb R^{2} \mid\text{ either } x,y \ge0 \text{ or }x,y \le0 \}.$$

To determine if it's closed under addition, let $u$ be $x_1, y_2$ and $v$ be $y_1, y_2$. It is closed if $u + v$ exists. I take $x_1 + x_2$ and $y_1 + y_2$, but I am unsure how to explain because of the conditions.

To determine if it's closed under scalar multiplication, I let $C$ be a negative scalar and $u$ be vector $x,y$. I was thinking if I choose a negative scalar, and choose the option that $x_1, y_2$ would have to be larger than $0$, then it would not be closed under scalar multiplication and thus not a subspace.

share|cite|improve this question
You omitted the parentheses around the ordered pairs; they’re required, so you should edit the question to include them. – Brian M. Scott Nov 2 '12 at 19:25
As a matter of fact, $(0,0)\notin H$, but every subspace contains $(0,0)$. – Hagen von Eitzen Nov 2 '12 at 19:37
Why $\,(0,0)\notin H\,$? In fact it is. – DonAntonio Nov 3 '12 at 11:12
up vote 2 down vote accepted

Your $H$ is closed under multiplication by scalars, but it is not closed under addition. If you consider $(2,1)$, $(-1,-2)$, then both are in $H$ but $(2,1)+(-1,-2)=(1,-1)$ is not in $H$.

share|cite|improve this answer

$(1,3)+(-3,-2)=(-2,1)$ so $H$ is not closed under addition and therefore is not a subspace of $\mathbb R^2$. (To help you visualise the set, it consists of all elements in the first and third quadrants on the Cartesian plane, including the axes.)

share|cite|improve this answer

The conditions are essentially that $(x,y)\in H$ if $(x,y)$ is in the first or third quadrant. To demonstrate that $H$ is not closed under addition, just show that the sum of some two vectors in the first and third quadrant is in the second or fourth quadrant.

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.