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Math Help - Vector Space/Linear Independece

  1. #1
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    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!!
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  2. #2
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    Quote Originally Posted by sabrepride View Post
    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 \mathbb{F}_3 are only \{0,1,2\} \!\!\!\pmod 3

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


    No, again: in \mathbb{F}_3\,,\,\,2\cdot 2=1 , since 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 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 v_!\,,\,v_2 are lin. ind. the above is possible iff 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!!
    .
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  3. #3
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    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 F\ne F_2 that implies a= 0. Then, of course, a+ b= 0+ b= 0 gives b= 0.
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  4. #4
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    Quote Originally Posted by tonio View Post
    .
    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.
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  5. #5
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    Quote Originally Posted by sabrepride View Post
    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 k=1\,,\,\,k=-1\Longrightarrow 1=-1\Longrightarrow 2=0\Longrightarrow p=2 , or at least q is a power of 2.

    Tonio
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