Question

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

2. $\mathit{W}\mathbf{=}\left\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z\right\}$

Short answer

$W=\left\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z\right\}$ is not a subset of role="math" localid="1660732804183" ${\mathrm{ℝ}}^3$.

Step-by-step solution

Consider the set.

A subset of the vector space ${\mathrm{ℝ}}^n$is called a (linear) subspace of 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.

$W=\left\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z\right\}$

Check for first condition.

The first condition is,

$W=\left\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z\right\}$

$\left[\begin{array}{c}0\\ 0\\ 0\end{array}\right]:0\le 0\le 0$

The first condition holds good.

Check for second condition.

The second condition is,

$$\begin{gathered} {\overrightarrow{w}}_1=\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z,{\overrightarrow{w}}_2=\left[\begin{array}{c}a\\ b\\ c\end{array}\right]:a\le b\le c \\ {\overrightarrow{w}}_1+{\overrightarrow{w}}_2=\left\{\left[\begin{array}{c}x+a\\ y+b\\ z+c\end{array}\right]:x+a\le y+b\le z+c\right\} \end{gathered}$$

The second condition holds good.

Check for third condition

The third condition is,

If $k>0$

$k{\overrightarrow{w}}_1=\left\{\left[\begin{array}{c}kx\\ ky\\ kz\end{array}\right]:kx\le ky\le kz\right\}$

If k<0

$k{\overrightarrow{w}}_1=\left\{\left[\begin{array}{c}kx\\ ky\\ kz\end{array}\right]:kx\le ky\le kz\right\}$

The third condition does not hold good.

Final answer.

$W=\left\{\left[\begin{array}{c}x\\ y\\ z\end{array}\right]:x\le y\le z\right\}$ is not a subset of ${\mathrm{ℝ}}^3$ .