# vector space

• Oct 2nd 2010, 01:56 PM
alexandrabel90
vector space
how do you show that for every set S, S ⊂ span S?

my working:

let S be the set of real numbers in n dimension.

then S = { x1,x2,...,xn l x1,x2,...,xn are real numbers}

then span S = a1x1+ a2x2+...+anxn

for all xi ∈ S, where 1<i<n

xi = a1x1+ a2x2+.+aixi+..+anxn iff all the coefficients are 0 except ai=1

thus xi ⊂ a1x1+ a2x2+.+aixi+..+anxn

can i prove this question like this? how should i define S since the proof requires that the statement is true for all S?

thanks
• Oct 2nd 2010, 01:59 PM
Ackbeet
So, $\displaystyle \text{span}(S)$ is the set of all linear combinations of vectors in $\displaystyle S.$ Correct? Let $\displaystyle s\in S$. Then $\displaystyle s=1\cdot s\in\text{span}(S),$ because it's a linear combination of vectors in $\displaystyle S.$

Doesn't this do it?
• Oct 2nd 2010, 02:00 PM
HappyJoe
I don't quite understand your proof.

But to show the statement for a general set S, notice that span S consists of all finite linear combinations of elements from S. In particular, take some element x in S. Then x=1x is a finite linear combination involving elements of S, and hence x is in span S.
• Oct 2nd 2010, 02:19 PM
alexandrabel90
thanks!! i was thinking that i need to define the vectors in S and then define the span of S.
• Oct 4th 2010, 01:50 AM
Ackbeet
You're very welcome for my contribution, whatever that was. Have a good one!