# Homework Help: Vector space question

1. Jul 30, 2010

### sakodo

1. The problem statement, all variables and given/known data
Let V be a vector space over the field K.

a) Let {$$W_{k}:\ 1\leq k \leq m$$} be m subspaces of V, and let W be the intersection of these m subspaces. Prove that W is a subspace of V.

b) Let S be any set of vectors in V, and let W be the intersection of all subspaces of V which contains S (that is, x E W if and only if x lies in every subspace which contains S). Prove that W is the set of finite linear combinations of vectors from S.

2. Relevant equations

3. The attempt at a solution
a) I got this part so I will skip this. Part b is where I am stuck at. Just assume W is a subspace of V.

b) From what I understand, the question wants me to prove that W=span of S. I seriously don't know what to do. I tried to prove that any vectors that are NOT the span of S cannot be in W, but I didn't know where to go from there.

From a book I read, b) is actually a theorem. It says "W is the smallest subspace of V that contains S" but unfortunately it doesn't show any proofs for it.

I feel like I have missed something. Any hints?

Any help would be appreciated.

2. Jul 30, 2010

### vela

Staff Emeritus
Show W is contained in span(S), and span(S) is contained in W. Then W=span(S).

3. Jul 30, 2010

### sakodo

I get what you mean. You are saying if A is a subset of B and B is a subset of A, then A=B.

Here is what I got so far:

let S={$$\lambda_{1},\lambda_{2}...\lambda_{n}$$}

then, $$A1\lambda_{1},A2\lambda_{2}...An\lambda_{n}$$ E W1,W2,...Wm. (Closure under multiplication by a scalar.)

and so $$A1\lambda_{1}+A2\lambda_{2}+...+An\lambda_{n}$$ E W1,W2,...Wm. (Closure under vector addition.)

And so, span(S) E W1,W2...Wm

Thus, span(S) is contained in W, as W is the intersection of W1,W2...Wm.

How do I prove that W is contained in span(S)?

4. Jul 31, 2010

### vela

Staff Emeritus
Consider the fact that span(S) is a subspace of V that contains S.

EDIT: I changed the wording in this post to say what I meant to say. Ignore what I had written here earlier.

Last edited: Jul 31, 2010
5. Jul 31, 2010

### sakodo

OMG I GOT IT.

Since the span of S is a subspace of V, and W is the intersection of the subspaces in V that contains S, then obviously W E span{S}.

Thus, span{S} E W and W E span{S} and so W=span{S}.

Therefore, W is the set of finite linear combinations of S. =)

Thanks so much man. You are awesome =D.