# Span of a Set

• Sep 29th 2006, 12:37 PM
TreeMoney
Span of a Set
The question is.......

Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

• Sep 29th 2006, 12:46 PM
CaptainBlack
Quote:

Originally Posted by TreeMoney
The question is.......

Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

Let x in Span(u,v), then there exists a, b in R, such that:

x=au+bv.

Now put a'=(a+b)/2 and b'=(a-b)/2, then x=a'(u+v) +b'(u-v) so x is in
Span(u+v, u-v).

A similar argument (left to the reader) will show that if y in Span(u+v, u-v)
then y is also in Span(u,v).

RonL
• Sep 29th 2006, 12:51 PM
ThePerfectHacker
Quote:

Originally Posted by TreeMoney
The question is.......

Let u and v be any vectors in R^n. Prove that the spans of {u,v} and {u+v, u-v} are equal.

I'm thinking if they u and v were inverses of each other then this would work, but i'm not sure.

We need to show that the sets,
S={au+bv|a,b in R and u,v in R^n}
T={a(u+v)+b(u-v)|a,b in R}
Are equal.

Equivalently, S is subset of T and T is subset of S.

In set T if a=1/2 and b=1/2 then, u is in S
In set T if a=1/2 and b=-1/2 then v is in S.
Thus, their linear combination is in T.
Thus, T contains S.

In set S if a=1 and b=1 then u+v is in T.
In set S if a=1 and b=-1 then u-v is in T.
Thus, their linear combination is in T.
Thus, S contains T.

Thus, T=S