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Problem

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 $\mathbb{R}^{3}$

Solution

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?

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3  
(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
4  
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 R-F Feb 18 '12 at 3:15
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1 Answer

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.

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