# Homework Help: Linear algebra proof check

1. Feb 15, 2012

### Dansuer

1. The problem statement, all variables and given/known data
Prove that if subspace W contain a set of vectors S, then W contain the span(S)

2. Relevant equations

3. The attempt at a solution
Let's take a vector $x\in span(S)$, i have to show $x\in W$ also. (*)
So since $x\in span(S)$ there are scalrs $c_1...c_n$ so that $x = c_1s_1 ...c_ns_n$ where $s_1...s_n$ are elements of S.
Let's take $s_1 = \frac{x}{c_1} - \frac{c_2}{c_1} - ...-\frac{c_n}{c_1}$ which is of course an elemtent of S.
Since $S \subseteq W$ s is an element of W also.
Since W is a vector space $c_1s_1 + c_2s_2 + ... + c_ns_n = x$ is still an element of W, so x is an element of W

I'd like a check, thanks :)

EDIT: I'm adding a part after the (*)
If x is the zero vector, then any space contains the zero vector and we are done. If x is not the zero vector then there are scalars $c_1...c_n$ where at least one is not zero, let that scalar be c_1, so that $x = c_1s_1 ...c_ns_n$ where $s_1...s_n$ . . .

Last edited: Feb 15, 2012
2. Feb 15, 2012

### Some Pig

If W contains S, since W is a subspace, any linear combinations of the vectors in S will also in W, hence W contains span(S).

3. Feb 15, 2012

### Dansuer

Thank you, and your proof is even much simpler and short than mine.
But i still would like to know if mine is correct.

4. Feb 15, 2012

### genericusrnme

I'd just use what Some Pig used too, if you're a subspace you contain the span of any combination of your vectors.

Regarding your proof, your last two lines are all you really need as such your proof is correct.

5. Feb 16, 2012

### Dansuer

Thanks a lot

6. Feb 16, 2012

### genericusrnme

No problem buddy!

7. Feb 16, 2012

### sunjin09

This is wrong (and useless) since s1 is already given, not what you specify, and x is a vector and c2/c1,etc. are scalars, the subtraction is undefined

8. Feb 17, 2012

### Dansuer

Yes i made a huge mistake in writing that, what i meant to say was $s_1 = \frac{x}{c_1} - \frac{c_2}{c_1}s_2 - ...-\frac{c_n}{c_1}s_n$