1. ## Vector Space/Linear Independece

So my professor told us this and I am trying to see how its true.

You have a vector space V. v1 and v2 are elements of V. V is over an arbitrary field F-sub-q (q not equal to two), so if it was F-sub-3 for example, its elements would go 0,1,2,0,1,2... so say you multiply 2 * 2 which would normally equal 4, here would equal 0.

v1 and v2 are non-collinear, so v1 does not equal lambda * v2 for for any lambda in F-sub-q.

He made the claim that v1+v2 is linearly independent of v1-v2.

I thought about the definition of that, so if x(v1+v2)+y(v1-v2)=0vector, then x,y MUST be 0. I am just having a hard time proving that.

Any and all help would be much appreciated!!

2. Originally Posted by sabrepride
So my professor told us this and I am trying to see how its true.

You have a vector space V. v1 and v2 are elements of V. V is over an arbitrary field F-sub-q (q not equal to two), so if it was F-sub-3 for example, its elements would go 0,1,2,0,1,2...

Uuh? No, the elements of $\displaystyle \mathbb{F}_3$ are only $\displaystyle \{0,1,2\} \!\!\!\pmod 3$

so say you multiply 2 * 2 which would normally equal 4, here would equal 0.

No, again: in $\displaystyle \mathbb{F}_3\,,\,\,2\cdot 2=1$ , since $\displaystyle 4=1\!\!\!\pmod 3$

v1 and v2 are non-collinear, so v1 does not equal lambda * v2 for for any lambda in F-sub-q.

He made the claim that v1+v2 is linearly independent of v1-v2.

In any vector space, two vectors are lin. ind. iff none of them is a scalar multiple of the other one, so assume $\displaystyle v_1+v_2=k(v_1-v_2)\,,\,\,k\in\mathbb{F}_q\Longrightarrow (k-1)v_1-(k+1)v_2=0$ . As $\displaystyle v_!\,,\,v_2$ are lin. ind. the above is possible iff $\displaystyle k=1\,,\,k=-1$ ...and this cannot be UNLESS the field's characteristic is 2...

Tonio

I thought about the definition of that, so if x(v1+v2)+y(v1-v2)=0vector, then x,y MUST be 0. I am just having a hard time proving that.

Any and all help would be much appreciated!!
.

3. A slightly different way: if a(x+ y)+ b(x- y)= 0, then (a+b)x+ (a-b)y= 0.

If x and y are independent (not collinear) then we must have a+ b= 0 and a- b= 0.

Adding those equations 2a= 0 and, as long as $\displaystyle F\ne F_2$ that implies a= 0. Then, of course, a+ b= 0+ b= 0 gives b= 0.

4. Originally Posted by tonio
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Could you clarify why that if k=1 and k=-1, p would have to equal 2.

Thanks so much for your help thus far, I think I get it now. I was thinking about proving this by contraction, but I like this more straghtforward way.

5. Originally Posted by sabrepride
Could you clarify why that if k=1 and k=-1, p would have to equal 2.

Thanks so much for your help thus far, I think I get it now. I was thinking about proving this by contraction, but I like this more straghtforward way.

Well, if both $\displaystyle k=1\,,\,\,k=-1\Longrightarrow 1=-1\Longrightarrow 2=0\Longrightarrow p=2$ , or at least q is a power of 2.

Tonio