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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?

  • 0
    So, It was kinda simple :( I'm sorry but didn't knew how to test it.2012-02-18

1 Answers 1

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  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.