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 .

Printable View

- Dec 1st 2011, 01:25 PMLotte1990Subspace of a vector space
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 . - Dec 1st 2011, 01:26 PMDrexel28Re: Subspace of a vector space
- Dec 1st 2011, 01:28 PMLotte1990Re: Subspace of a vector space
- Dec 1st 2011, 01:43 PMDrexel28Re: Subspace of a vector space
Well, I'll give you a hint. As I said, the problem is that being a subgroup of does not imply that is a subspace. Said more concretely, we have that implies DOES NOT imply that for all . To see this, look at my example. For all one has that so that is closed under addition. But, look at where the problem comes in, is closed under arbitrary real number multiplication?

- Dec 1st 2011, 02:16 PMHartlwRe: Subspace of a vector space