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$$\{(x,y,z)|2x+9y=0, 8x-5z=0\}$$

I solved these as simultaneous equations giving me the equation $36y + 5z = 0$ or $y=-\frac{5}{36}z$, which I can write as

$$ \begin{bmatrix} -\frac{5}{36}z \\ z \\ \end{bmatrix} = 0.$$

Substituting $0$ for $z$ satisfies this equation. And it is closed multiplication and addition so this the original set is a subspace of $\mathbb{R}^3$. Is that method correct or am I completely off with this?

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You forgot to write $x$ in terms of $z$. –  lhf Mar 19 '12 at 11:29
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2 Answers

up vote 3 down vote accepted

After you solve $y = -\frac{5}{36}z$, you should substitute into the first equation to find a relationship with $x$:

$$\begin{align*} 2x + 9\left(-\frac{5}{36}z\right) &= 0 \\ 2x - \frac{5}{4}z &= 0 \\ x &= \frac{5}{8}z \end{align*}$$

This implies all vectors in the set can be written in the form: $$ \left[\begin{array}{c} x \\ y \\ z \\ \end{array}\right] = \left[\begin{array}{c} \frac{5}{8}z \\ -\frac{5}{36}z \\ z \\ \end{array}\right] = \left[\begin{array}{c} \frac{5}{8} \\ -\frac{5}{36} \\ 1 \\ \end{array}\right]z$$

And this implies the set is a vector subspace of $\mathbb{R}^3$, since it can be written as the span of some number of vectors of $\mathbb{R}^3$.

There are other ways to analyze the set, but hopefully this method is closest to what you have already tried.

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Let $V=\{(x,y,z)|2x+9y=0, 8x-5z=0\}$. To check that $V$ is a linear subspace of $\mathbb{R}^3$ we need to check $3$ things:

  • $0=(0,0,0)\in V$,
  • If $v\in V$ and $\lambda\in\mathbb{R}$, then $\lambda\cdot v \in V$, and
  • If $v,w\in V$, then $v+w\in V$.

Let's show that these three conditions are satisfied:

  • $0=(0,0,0)\in V$: Indeed $2\cdot 0 +9\cdot 0 = 0$ and $8\cdot 0 - 5\cdot 0 =0$, so $(0,0,0)\in V$.

  • If $v\in V$ and $\lambda\in\mathbb{R}$, then $\lambda\cdot v \in V$: suppose $v=(x_0,y_0,z_0)$ and $v\in V$. Then, $2x_0+9y_0=0$ and $8x_0-5z_0=0$. Thus, $$2(\lambda x_0)+9(\lambda y_0) = \lambda (2x_0+9y_0)=\lambda(0)=0$$ and $$8(\lambda x_0)-5(\lambda z_0)= \lambda(8x_0-5z_0)=\lambda (0)=0.$$ Hence, $\lambda v = (\lambda x_0, \lambda y_0, \lambda z_0) \in V$ as well.

  • If $v,w\in V$, then $v+w\in V$: let $v=(x_0,y_0,z_0)$ and $w=(x_1,y_1,z_1)$, and suppose $v,w\in V$. Then, $2x_0+9y_0=0$ and $8x_0-5z_0=0$, and $2x_1+9y_1=0$ and $8x_1-5z_1=0$. Hence: $$2(x_0+x_1)+9(y_0+y_1)= (2x_0+9y_0)+(2x_1+9y_1)=0+0=0$$ and $$8(x_0+y_0)-5(z_0+z_1)=(8x_0-5z_0)+(8x_1-5z_1)=0+0=0.$$ Hence $v+w\in V$ as well.

All three conditions are verified and therefore $V$ is a linear subspace of $\mathbb{R}^3$.

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