Thread: 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.

I'm not sure how to go about this problem. TIA

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.

I'm not sure how to go about this problem. TIA

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

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.

I'm not sure how to go about this problem. TIA

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