Question

Let S be a non empty finite subset of a Vector Space V over the field F. And let W be the set of all linear combinations of S.
Can we always say L(S)=W
where L(S) refers to the linear span of the set S

Answer 1

That's the definition of the span of S.

Answer 2

Actually I am currently reading a book which defines L(S) as the smallest subspace containing S, and I felt that then obviously L(S)=W, but the book never states that

Answer 3

So I thought perhaps I was doing something wrong

Answer 4

Ok, that makes sense. The answer is yes and the proof isn't too bad.

W is a subspace that contains S hence L(S) is a subset of W. Now show that L(S) must be a subset of W.

Answer 5

I have the proof even, in the same book, but the confusing part is the book never states is directly

Answer 6

And another thing that confuses me is the following.
The book goes on to prove that if S is a subset of T, then L(S) is a subset of L(T).
While doing that

Answer 7

They are first saying
let S={a_!,a_2...a_n}........

Answer 8

Then they are saying, let M = {c_1.a_1,c_2.a_@..........c_n.a_n}

Answer 9

that @ should have actually been 2

Answer 10

Next, I reasoned that since S is a subset of T, S is {a_!,a_2...a_n} belongs to T too,
Now since L(T) is the set of all linear combination of T, {c_1.a_1,c_2.a_@..........c_n.a_n} belongs to L(T)

Answer 11

I'm going to stop there.

Answer 12

Are you getting to a question about the proof in the book? Or are you trying to construct your own proof?

If the later, I have a short, neat proof that doesn't involve all of the ugliness of explicitly writing down linear combinations of vectors.

Answer 13

I'll give anyway. If S is a subset of T, every subspace that contains T also contains S. Hence L(S) is a subset of any subspace that contains T. Therefore L(S) is a subset (maybe not a strict one, but still a subset) of L(T).

Answer 14

Hey, I am sorry about that long quietude, I was actually trying to type my answer, by the webpage turned down.

Answer 15

I think the site is back up to speed now. It was very slow for me as well, but is now behaving well.

Answer 16

But that link was a live link, it would help both of us to discuss this faster. If you don't have any problem in visiting that, then I would suggest that.

Answer 17

Thanks a lot for your help, at the end I couldn't reply there on that sheet too, because some problem arose there too :(
Thanks again.

Answer 18

Can you please take a look at the question posted on http://openstudy.com/#/updates/4f0455ece4b075b56651b5d6