# Thread: exams soon....Prove that {u+v,u-v} is a basis for S

1. ## exams soon....Prove that {u+v,u-v} is a basis for S

Dear Smart people,
Let u and v be linearly independent vectors in a vector space V.
Let S=Span{u,v}
Prove that {u+v,u-v} is a basis for S
thanks!

2. ## Re: exams soon....Prove that {u+v,u-v} is a basis for S

Originally Posted by n22
Let u and v be linearly independent vectors in a vector space V.
Let S=Span{u,v}
Prove that {u+v,u-v} is a basis for S.
You must show that $\displaystyle u+v~\&~u-v$ are linearly independent vectors.

You must show that if $\displaystyle w\in S$ then $\displaystyle \exists~\alpha~\&~\beta$ such that $\displaystyle w=\alpha(u+v)+\beta(u-v)$

3. ## Re: exams soon....Prove that {u+v,u-v} is a basis for S

First, since u and v are linearly independent vectors that span S, they are a basis for S.
To show that {u- v, u+ v} is a basis for S, you must show that they also are independent and span S.

So you must show that if a(u- v)+ b(u+ v)= 0 for numbers, a and b, then a= 0 and b= 0. You can do that by combing multiples of u and combining multiples of v and using the fact that u and v are independent.

To show that u- v and u+ v span S you must show that any vector in S can be written as a linear combination of u- v and u+ v. Of course, because u and v span S, any vector in S can be written as au+ bv.