Does a subspace containing a set of vectors also contain their span?

Dansuer
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Homework Statement


Prove that if subspace W contain a set of vectors S, then W contain the span(S)

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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 . . .
 
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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).
 
Thank you, and your proof is even much simpler and short than mine.
But i still would like to know if mine is correct.
 
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.
 
Thanks a lot
 
No problem buddy!
 
Dansuer said:

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.


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
 
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
 
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