# Linear Independence

Printable View

• Sep 30th 2006, 05:42 AM
TreeMoney
Linear Independence
This is like the other question I posted. I get so confused when they start talking about u + v and all these variables. Give me #'s and I'm good. :) But Variables throw me off for some reason. Anyway, I can't figure this out either

Let u and v be distinct vectors in R^n. Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.

TIA!!!!
• Sep 30th 2006, 08:54 AM
CaptainBlack
Quote:

Originally Posted by TreeMoney
This is like the other question I posted. I get so confused when they start talking about u + v and all these variables. Give me #'s and I'm good. :) But Variables throw me off for some reason. Anyway, I can't figure this out either

Let u and v be distinct vectors in R^n. Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.

TIA!!!!

Suppose {u,v} are linearly independent, and that {u+v,u-v} are not,

Then there exist a, b in R, at least one of which !=0, such that:

a(u+v) + b(u-v)=0,

so:

(a+b)u + (a-b)v=0

but now at least one of (a+b) and (a-b) !=0, so we conclude that
u and v are linearly dependent - a contradiction.

So we conclude that if {u.v} are linearly independent then so are {u+v,u-v}.

The second half of the proof proceeds like the above but starting from
the assumption that {u+v,u-v} are linearly independent and that [u,v}
are not.

RonL
• Sep 30th 2006, 05:36 PM
ThePerfectHacker
Quote:

Originally Posted by TreeMoney
Prove that {u,v} is linearly independent if and only if {u+v, u-v} is linearly independant.

We now that if,
Cu+Kv=0 then, C=K=0
We need to show if,
C(u+v)+K(u-v)=0 then, C=K=0
That means, (scalar multiplication is commutatiove in vector space)
Cu+Cv+Ku-Kv=0
Thus,
(C+K)u+(C-K)v=0
By our initial assumption of linear independence if {u,v}
C+K=0
C-K=0
The only solutions is 0.

Now, for the converse.
If, C(u+v)+K(u-v)=0
Then, C=K=0
Thus, we have,
Cu+Kv=0 we need to show C=K=0
Do the following trick, (skipping the distributative commutative stuff)
(C+K)(u+v)+(C-K)(u-v)=0
Thus,
C+K=0
C-K=0
Thus,
C=K=0