# Linear independance

1. Jul 26, 2012

### JeeebeZ

1. The problem statement, all variables and given/known data

If vectors u v & w ARE linearly independent are:

a) u+v, u+w, v+w
b) u-v, u-w, v-w

The attempt at a solution

I don't really know where to start with this. It isn't homework. It is just one of the questions out of my textbook(ch 2.3 #43 linear algebra by poole 3rd ed).

So the answers are a) yes b) no. But the question says to explain why and the back of the book just says yes and no. I don't understand why either would be different.

I guessed and said the first one would be independent. But guessing doesn't really do anything in the real world. other then gives you a small percentage to be right.

Any help would be muchly appreciated

2. Jul 26, 2012

### LCKurtz

Start with the definition. Can you or can you not find constants $A,B,C$ not all zero such that$$A(u+v) + B(u+w) + C(v+w)=0$$

3. Jul 26, 2012

### JeeebeZ

$$u(A+B) + v(A+C) + w(B+C) = 0$$

$$u(A+B) + v(-A+C) + w(-B-C) = 0$$
A=1, B=-1, C=1.
1+-1=0, -1+1=0, --1-1=0

So.. By inspection.. I can make the second one a 0 without zero's.. So that would mean. They are now linearly dependent. But how would I calculate that? I don't know how to put a proof vector in to a real formula.

4. Jul 26, 2012

### Ray Vickson

In the first case, you need A+B=0, A+C=0, B+C=0, because these are the coefficients of u, v and w, and u,v,w are linearly independent. The only solution is A = B = 0, so that means the original set {u+v,u+w,v+w} is linearly independent as well.

In the second case you need A+B=0, -A+C=0, -B-C=0. There is a nonzero solution, so the set {u-v,u-w,v-w} is linearly dependent.

RGV