1. ## vector subspace

If V={(x_1, x_2) ∈ R2|x_1≤0, x_2≤0} Is V a linear subspace of R2?

Sol: I wrote, it's not a subspace since the vector -v= [-1,-1] belongs to the set but v does not belong to it. Am I right? Help!

2. Originally Posted by Taurus3
Am I right?

Yes, you are.

Fernando Revilla

3. Just to elaborate a bit:

$\displaystyle V$ is not an $\displaystyle \mathbb{R}$-subspace of $\displaystyle \mathbb{R}^2$ because it is not closed under scalar multiplication by elements from the field $\displaystyle \mathbb{R}$ (which is basically what your example is saying).

That is, dealing with vector spaces over some field $\displaystyle F$, one of the conditions for $\displaystyle V\subseteq W$ to actually be a subspace of $\displaystyle W$ is that, for any $\displaystyle \alpha \in \mathbb{R},v\in V$, we must have $\displaystyle \alpha v\in V$.

This is not the case in your example. As you pointed out, $\displaystyle -1\in \mathbb{R}$ and $\displaystyle <-1,-1>\in V$, but $\displaystyle -1v=<1,1> \notin V$.