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


Determine whether the indicated subset is a subspace of the given euclidean space:

$ \{[x,y,z]\ |\ x,y,z \in \mathbb{R} $ and $z=3x+2\}$ in $\mathbb{R}^{3}$


By definition, in order for a subset to be a subspace 3 conditions must be occur:

  1. To pass by the origin To contain the origin.
  2. To be closed under addition.
  3. To be closed under scalar multiplication.

So I try to solve the exercise by this way:

$1.$ The origin $(0,0,0) \in \mathbb{W} $

$2.$ Let $\vec u$ and $\vec v \in \mathbb{W} $. We have

$$ \begin{cases} 3u_1 + 2 - u_3 = 0 \\ 3v_1 + 2 - v_3 = 0 \\ \end{cases} $$ The sum is $ 6(u_1 + v_1) + 4 - (u_3+v_3) = 0 $ (which $\in \mathbb{W} $)

$3.$ Let $\vec u$ $ \in \mathbb{W} $ and $\ r$ $ \in \mathbb{R} $. We have

$r(3u_1) + r(2) - r(u_3) = 0 \\$

Which, also, $ \in \mathbb{W} $

So, why is the book's answer: It isn't a subspace?

share|cite|improve this question
(1) Sets don't "pass by the origin" (or through the origin), they either contain it or not. (2) Did you really mean $x,y,z$ to be in $\mathbb{R}^2$ and not in $\mathbb{R}$. (3) Check again: does $(0,0,0)$ really lie in the set? Is the third coordinate equal to three times the first plus 2? – Arturo Magidin Feb 18 '12 at 3:09
You seem to be confused about what it means to "be in" a subspace. For example, $(0,0,0)$ is not in your subspace, because if you substitute $x=y=z=0$ into $z=3x+2$ you do not get a true statement. – Jason DeVito Feb 18 '12 at 3:10
So, It was kinda simple :( I'm sorry but didn't knew how to test it. – Randolf Rincón Fadul Feb 18 '12 at 3:15
up vote 3 down vote accepted
  1. The origin is not in the set. In order to be in the set, you need $z=3x+2$. For $(0,0,0)$, $x=z=0$, but $0\neq 3(0)+2$.

  2. The set is not closed under sums. In order for $$(u_1,u_2,u_3)+(v_1,v_2,v_3)=(u_1+v_1,u_2+v_2,u_3+v_3)$$ to be in the set, you need $u_3+v_3 = u_1+v_1 + 2$. Your sum of equations proves nothing, and saying that the equation (which is false) "is in $\mathbb{W}$" is false; equations (or values) are not in $\mathbb{W}$, only vectors can be in $\mathbb{W}$.

    For an explicit example, note that $(0,0,2)$ and $(1,0,5)$ are both in $\mathbb{W}$. The sum is $(1,0,7)$, but $(1,0,7)$ is not in $\mathbb{W}$, because $7\neq 3(1)+2$.

  3. The set is not closed under scalar multiplication. In order for $$r(u_1,u_2,u_3) = (ru_1,ru_2,ru_3)$$ to be in $\mathbb{W}$, you need $ru_3 = 3(ru_1)+2$. From the assumption that $u_3=3u_1+2$ you cannot conclude that $ru_3 = 3(ru_1)+2$. In fact, this is false for any $r\neq 1$. For an explicit example, note that $(0,0,2)$ is in $\mathbb{W}$, but $2(0,0,2) = (0,0,4)$ is not, because $4\neq 3(0)+2$.

So the reason it is not a subspace is that it fails every single one of the three requirements.

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.