0
$\begingroup$

I wanted to determine if $W$ is a subspace of $\mathbb{R}^3$:

$W=\{(x,y,z)\in\mathbb{R}^3:x\leq y\leq z\}$.

I believe that it is a subspace of $\mathbb{R}^3$ since I think it satisfies all the conditions (contains the zero vector, is closed under addition and scalar multiplication), is this assumption correct?

Thank You!

$\endgroup$
3
  • 2
    $\begingroup$ It's not closed under scalar mutliplication. $\endgroup$
    – user296602
    Jul 5, 2016 at 5:11
  • $\begingroup$ Is it because this inequality limits you only to numbers less than or equal to zero, meaning that if you multiply it, it will surpass the given space? $\endgroup$
    – Irina
    Jul 5, 2016 at 5:17
  • $\begingroup$ You have to allow for multiplication by negative scalars as well as non-negative scalars. $\endgroup$ Jul 5, 2016 at 6:11

2 Answers 2

3
$\begingroup$

If $(x,y,z)\in\mathbb{R}^3$ satisfies the inequality, multiplying it by the scalar $-1\in\mathbb{R}$ will reverse the inequality.

$\endgroup$
0
$\begingroup$

$\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}1\\2\\3\end{bmatrix}$ is an element of $W$ since $1\leq 2\leq 3$ (it satisfies $x\leq y\leq x$)

Is $-1\cdot \begin{bmatrix}1\\2\\3\end{bmatrix}$ an element of $W$?

Is it true that $-1\leq -2\leq -3$?

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .