Question

Which of the sets W in Exercises 1 through 3 are subsets of${\mathbf{ℝ}}^{\mathbf{3}}$ ?

3. $\left\{W=\left[\begin{array}{c}x+2y+3z\\ 4X+5y+6z\\ 7x+8y+9z\end{array}\right]:x,y,zarbitraryconstants\right\}$

Short answer

$W=\left\{\left[\begin{array}{c}x+2y+3z\\ 4x+5y+6z\\ 7X+8y+9z\end{array}\right]:x,y,z\mathrm{arbitrary}\mathrm{constants}\right\}$ is a subset of ${\mathrm{ℝ}}^3$ .

Step-by-step solution

Consider the set.

A subset W of the vector space ${\mathrm{ℝ}}^n$is called a (linear) subspace of ${\mathrm{ℝ}}^n$if it has the following three properties:

a. W contains the zero vector in ${\mathrm{ℝ}}^n$.

b. W is closed under addition: If ${\overrightarrow{w}}_1$and ${\overrightarrow{w}}_2$are both in W, then so is ${\overrightarrow{w}}_1+{\overrightarrow{w}}_2$.

c. W is closed under scalar multiplication: If $\overrightarrow{w}$is in W and k is an arbitrary scalar, then $k\overrightarrow{w}$is in W.

The set W should satisfy all the above conditions.

$W=\left\{\left[\begin{array}{c}x+2y+3z\\ 4x+5y+6z\\ 7X+8y+9z\end{array}\right]:x,y,z\mathrm{arbitrary}\mathrm{constants}\right\}$

Check for first condition.

The first condition is,

$W=\left\{\left[\begin{array}{c}x+2y+3z\\ 4x+5y+6z\\ 7X+8y+9z\end{array}\right]:x,y,z\mathrm{arbitrary}\mathrm{constants}\right\}$

Substitute, $x=1,y=-2,z=1$

$$\begin{gathered} x+2y+3z=\left(1\right)+2\left(-2\right)+3\left(1\right)=0 \\ \Rightarrow 0\in W \\ 4x+5y+6z=4\left(1\right)+5\left(-2\right)+6\left(1\right)=0 \\ \Rightarrow 0\in W \\ 7x+8y+9z=7\left(1\right)+8\left(-2\right)+9\left(1\right)=0 \\ \Rightarrow 0\in W \\ \therefore \left(0,0,0\right)\in W \end{gathered}$$

The first condition holds good.

Check for second condition.

The second condition is,

$$\begin{gathered} {\overrightarrow{w}}_1=\left[\begin{array}{c}{x}_1+2y_1+3z_1\\ 4x_1+5y_1+6z_1\\ 7x_1+8y_1+9z_1\end{array}\right],{\overrightarrow{w}}_2=\left[\begin{array}{c}{x}_2+2y_2+3z_2\\ 4x_2+5y_2+6z_2\\ 7x_2+8y_2+9z_2\end{array}\right] \\ {\overrightarrow{w}}_1+{\overrightarrow{w}}_2=\left(\begin{array}{c}{x}_1+x_2\\ 4x_1+4x_2\\ 7x_1+7x_2\end{array}\right)+\left(\begin{array}{c}2y_1+2y_2\\ 5y_1+5y_2\\ 8y_1+8y_2\end{array}\right)+\left(\begin{array}{c}3z_1+3z_2\\ 6z_1+6z_2\\ 9z_1+9z_2\end{array}\right) \end{gathered}$$

${\overrightarrow{w}}_1+{\overrightarrow{w}}_2\in W$

The second condition holds good.

Check for third condition

The third condition is,

If $k=\mathrm{ℝ}$

$\overrightarrow{w}=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]$

$k{\overrightarrow{w}}_1=\left[\begin{array}{c}kx\\ ky\\ kz\end{array}\right]:kx\left[\begin{array}{c}1\\ 4\\ 7\end{array}\right]+ky\left[\begin{array}{c}2\\ 5\\ 8\end{array}\right]+kz\left[\begin{array}{c}3\\ 6\\ 9\end{array}\right]$

The third condition also holds good.

Final answer.

$W=\left\{\left[\begin{array}{c}x+2y+3z\\ 4x+5y+6z\\ 7X+8y+9z\end{array}\right]:x,y,zarbitrarycons\tan ts\right\}$ is a subset of role="math" localid="1660736135692" ${\mathrm{ℝ}}^3$ .