# Proof of span(S) is a subspace:a question

• Apr 6th 2011, 05:42 AM
x3bnm
Proof of span(S) is a subspace:a question
We know the definition of subspace which is:

If $\displaystyle W$ is a set of one or more vectors from a vector space $\displaystyle V$, then $\displaystyle W$ is a subsace of $\displaystyle V$ if and only if the following conditions hold:

(a) If $\displaystyle u$ and $\displaystyle v$ are vectors in $\displaystyle W$, then $\displaystyle u + v$ is in $\displaystyle W$.
(b) If $\displaystyle k$ is any scalar and $\displaystyle u$ is any vector in $\displaystyle W$, then $\displaystyle ku$ is in $\displaystyle W$.

Also the definition of $\displaystyle span(S)$ is:
If $\displaystyle S = \{v_1, v_2, ..., v_r\}$ is a set of vectors in a vector space $\displaystyle V$, then the subspace $\displaystyle W$ of $\displaystyle V$ consisting of all linear combinations of the vectors in $\displaystyle S$ is called the space spanned by $\displaystyle v_1, v_2, ...,v_r$, and we say that the vectors $\displaystyle v_1, v_2, ..., v_r$ span $\displaystyle W$. To indicate that $\displaystyle W$ is the space spanned by the vectors in the set $\displaystyle S = \{v_1, v_2,..., v_r\}$, we write:

$\displaystyle W = span(S)$ or $\displaystyle W = span\{v_1, v_2,..., v_r\}$

Now the problem I'm having is how is it possible that $\displaystyle span(S)$ is a subspace? I can't make a connection with the definition and the proof.

Is it possible to explain the proof of why $\displaystyle span(S)$ is a subspace?
• Apr 6th 2011, 06:12 AM
Deveno
what does an element of span(S) look like? more pointedly, how do you form a "linear combination of the vj" what does that MEAN?

when you have hit upon that, you will see at once that two such linear combinations may be combined, and that BY THEIR LINEARITY, combined into a single linear combination.

here is a two-dimensional example: suppose S = {u,v}. we may regard an element of span(S) as being of the form a1u + a2v. another such element might be: b1u + b2v.

what could be more natural than (a1u + a2v) + (b1u + b2v) = (a1+b1)u + (a2+b2)v. is this not a linear combination of u and v, as well?

extend this idea to a scalar product k(a1u + a2v).

it should be clear that if S is a non-empty set, that span(S) has "one or more vectors" in it. convince yourself that for u in S, ku in span(S) implies that the 0-vector is in span(S).

can you see that -u has to be in span(S) as well? what further convincing might you need to see that span(S) has all the properties of a vector space in and of itself, and being a subset, is thus a sub-space?
• Apr 6th 2011, 06:18 AM
HallsofIvy
The span, W, of a set of vectors $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}$ is, as you say, the set of all possible linear combinations of those vectors. Now, suppose u and v are vectors in W. Then $\displaystyle u= a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n$ for some numbers $\displaystyle a_1, a_2, \cdot\cdot\cdot, a_n$. Also, $\displaystyle v= b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n$.

Then, $\displaystyle u+ v= (a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)+ (b_1v_1+ b_2v_2+ \cdot\cdot\cdot+ b_nv_n)= (a_1+ b_1)v_1+ (a_2+ b_2)v_2+ \cdot\cdot\cdot+ (a_n+ b_n)v_n$ which is again a linear combination of $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}$ and so in W. Similarly, $\displaystyle ku= k(a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_nv_n)= ka_1v_1+ ka_2v_2+ \cdot\cdot\cdot+ ka_nv_n$ a linear combination of $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, v_n\}$ and so in W.
• Apr 6th 2011, 04:31 PM
x3bnm
I don't how I missed such a simple thing. It makes a lot of sense now. Thanks HallsofIvy and Deveno both for your explanations and extraordinary brains.(Happy)