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About LinkBacksThe 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. ConsiderOriginally Posted by Lotte1990
. Said differently, (if you had changed
to
) being a subgroup of a vector space's underlying abelian group is not equivalent to being a subspace of the vector space.
What do you mean? Could you be more specific? Could you define two specific vectors for w1 and w2? Or is this not possible?This is not true. Consider
I need some help in understanding this...
Well, I'll give you a hint. As I said, the problem is thatbeing a subgroup of
does not imply 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
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