# spanning

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• Oct 2nd 2010, 11:40 PM
alexandrabel90
spanning
how do i prove that every subspace is a spanning set for itself?

sorry im very weak at proofing.

my working:

let S be the subspace.
so we need to show that S = span(S)

we shall show that S⊂span S ( shown in previous post) and span S ⊂S

to show that spanS ⊂ S,

lets assume that S is in n dimension where xi∈ S where 1<i<n

span S =a1x1+...+anxn ⊂ S as S being a subspace obeys the laws of scalar multiplication and addition.

is this done like this?
• Oct 3rd 2010, 12:08 AM
HappyJoe
Your wording seems a bit off to me, but your idea seems quite right.
• Oct 3rd 2010, 01:37 AM
alexandrabel90
how should i change my explanation then?

thanksss!
• Oct 3rd 2010, 01:53 AM
HappyJoe
You want to show that when S is a subspace, we have that span S is contained in S. We don't really need to know anything about the dimension of S.

Take an element y in span S. Since span S consists of finite linear combinations of elements from S, we can write y = a_1x_1+...+a_nx_n for some elements x_i in S, and some scalars a_i (note: this n need not coincide with the dimension of S). And then, as you say, since S is closed under scaling and under vector addition, then a_1x_1+...+a_nx_n will also be in S, whereas y will be in S. So all elements of span S are contained in S - hence span S is a subset of S.
• Oct 3rd 2010, 03:41 AM
HallsofIvy
If v is in S then v= 1v. QED

(You posted this same question 10 hours earlier and got this same answer from Ackbeet, thanked him and said you understood. What went wrong?)
• Oct 3rd 2010, 04:20 AM
alexandrabel90
that previous question that i think you are referring to is for S ⊂ span(S) which i stated in my question here that i know how to solve that. now im trying span(S) ⊂ S