# subset of a vector space

• Oct 10th 2009, 04:54 PM
alexandrabel90
subset of a vector space
how do you prove this theorem?

theorem:
let S be a subset of a vector space V
(a) Then span S is a subspace of V which contains S
(b) if W is a subspace of V containing S, then span s ⊆ W

i would think that,

assume v, w ∈ span S, where
v= a1va + a2v2+...+ amvm and w=b1w1+b2w2+...bmwm
then

v+w=a1v1+..+amvm+b1w1+...bmwm.
thus span S is a linear combination of the vectors in S.

does that show that it is a subspace of V? it seems to me that i have just proven that span S is a vector space....

then how do i do the second part?

thanks!
• Oct 10th 2009, 07:28 PM
aman_cc
Quote:

Originally Posted by alexandrabel90
how do you prove this theorem?

theorem:
let S be a subset of a vector space V
(a) Then span S is a subspace of V which contains S
(b) if W is a subspace of V containing S, then span s ⊆ W

i would think that,

assume v, w ∈ span S, where
v= a1va + a2v2+...+ amvm and w=b1w1+b2w2+...bmwm
then

v+w=a1v1+..+amvm+b1w1+...bmwm.
thus span S is a linear combination of the vectors in S.

does that show that it is a subspace of V? it seems to me that i have just proven that span S is a vector space....

then how do i do the second part?

thanks!

Hi
For part 1. Please show that if $x\in$span(S) => $x\in V$
You will be done (This will prove span is a subset)

You have already done it in part one. Just replace V with W - the question is similar to part 1 above. Here you jut have to prove span S is a subset of W. A stronger inference is that it is a sub-space of W as well
• Oct 11th 2009, 01:56 AM
alexandrabel90
sorry!

could you explain further becos i cant understand what you mean..
• Oct 11th 2009, 03:00 AM
Failure
Quote:

Originally Posted by alexandrabel90
how do you prove this theorem?

theorem:
let S be a subset of a vector space V
(a) Then span S is a subspace of V which contains S
(b) if W is a subspace of V containing S, then span s ⊆ W

i would think that,

assume v, w ∈ span S, where
v= a1va + a2v2+...+ amvm and w=b1w1+b2w2+...bmwm
then

v+w=a1v1+..+amvm+b1w1+...bmwm.
thus span S is a linear combination of the vectors in S.

Be more careful here: the span of S is not a linear combination of the vectors of S. Instead it is the set of all linear combinations of vectors from S.

Quote:

does that show that it is a subspace of V? it seems to me that i have just proven that span S is a vector space....
First, before attempting to prove any proposition you have to be very clear about how the terms in that proposition have been defined.
Because the span of a set S with respect to a vector space V can be defined in different ways, you may or may not have been successful in your attempt. (I can think of at least three ways to define the span of S: first, as the set of all linear combinations of S; second, as the smallest subspace of V containing S; and, third, as the intersection of all subspaces of V containing S.)
Basically, you need to show, basing yourself on the definition of span S that you have been given (somwhere), that that set of vectors satisfies all the axioms of a vector space.

Quote:

then how do i do the second part?

thanks!
Surely, if S is a subset of W and you are forming a linear combination of elements of S (and thus of W), you get another element of W, because W is a vector (sub-)space: vector (sub-)spaces are, by definition, closed under the operation of forming linear combinations of some of their elements.