The following statement is false. Could someone give a counterexample?
If V is a real vector space and W is a (non-empty) subset of V such that for all vectors
w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .
The following statement is false. Could someone give a counterexample?
If V is a real vector space and W is a (non-empty) subset of V such that for all vectors
w1,w2 Є W it holds that also w1 + w2 Є W, then W is a subspace of V .
This is not true. Consider $\displaystyle \mathbb{Z}\subseteq\mathbb{R}$. Said differently, (if you had changed $\displaystyle w_1+w_2$ to $\displaystyle w_1-w_2$) being a subgroup of a vector space's underlying abelian group is not equivalent to being a subspace of the vector space.
Well, I'll give you a hint. As I said, the problem is that $\displaystyle W$ being a subgroup of $\displaystyle V$ does not imply that $\displaystyle W$ is a subspace. Said more concretely, we have that $\displaystyle w_1,w_2\in W$ implies $\displaystyle w_1+w_2\in W$ DOES NOT imply that $\displaystyle \alpha w_1\in W$ for all $\displaystyle \alpha\in\mathbb{R}$. To see this, look at my example. For all $\displaystyle x,y\in\mathbb{Z}$ one has that $\displaystyle x+y\in\mathbb{Z}$ so that $\displaystyle \mathbb{Z}$ is closed under addition. But, look at where the problem comes in, is $\displaystyle \mathbb{Z}$ closed under arbitrary real number multiplication?