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If (a,b,c) is a Basis of R3 , Does (a+b,b+c,c+a) also a Basis of R3

  1. Jul 17, 2010 #1
    1. The problem statement, all variables and given/known data
    Well the same as the subject..
    (a,b,c) is a Basis of [tex]R^{3}[/tex]
    does (a+b , b+c , c+a) Basis to [tex]R^{3}[/tex]

    I have another question ..
    is (a-b , b-c , c-a) Basis to [tex]R^{3}[/tex]
    This is know is not true because if I use e1, e2 , e3 I got a error line.

    2. Relevant equations
    start of liner algebra one.

    3. The attempt at a solution

    I assume that this is true..
    I think something like that.. Because (a,b,c)_ base of [tex]R^{3}[/tex] its means that I can get to a step matrix (Row echelon form)
    and the same with (b,c,a) and then I can just add them and get to any soultion that I want... But maybe I'm wrong?
    Thank you.

    Can I write base instead of Basis ?
  2. jcsd
  3. Jul 17, 2010 #2


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    Gold Member

    Attacking questions like this is one of the reasons you learn about things like "row spaces", "Gaussian elimination", and "coordinates relative to a basis".
  4. Jul 17, 2010 #3


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    I would be inclined to just use the fact that three vectors form a basis for a three dimensional space if and only if they are independent- and use the definition of "independent".
    a+ b, b+ c, and a+ c form a basis for R3 if and only if the only values of x, y, and z that make x(a+ b)+ y(b+ c)+ z(a+ c)= 0 are x= y= z= 0. Re write that as expressions in x, y, and z times "a", "b", and "c" separately and use the fact that a, b, and c are independent.
  5. Jul 17, 2010 #4


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    Presumably by (a,b,c) is a basis for R^3, you mean that the vectors (1,0,0), (0,1,0) and (0,0,1) form a basis for R^3, it seems that the question is asking you if (1,0,1), (1,1,0) and (0,1,1) form a basis for R^3. The question is then can you find three non-zero real numbers such that:
    If the answers is yes then the vectors don't form as basis, if the only solution is a_{i}=0, then they do form a basis.
  6. Jul 17, 2010 #5
    It can be shown, through elemental row procedures (on a 3x3 basis matrix) that the second basis: (a+b,b+c,c+a), could be reduces to the first one: (a,b,c).
    thus it is a basis of R3.

    I've checked, it can be done.

    Try it for yourself :)

    *P.S: the point of n vectors forming a basis in R^n space is that they aren't linearly dependent !
    Last edited: Jul 17, 2010
  7. Jul 17, 2010 #6


    Staff: Mentor

    No, you can't assume that a, b, and c are any specific vectors - only that they form a basis for R3.
    To the OP, to answer your question - you can have one basis or two bases. That's how I remember it working.
  8. Jul 17, 2010 #7
    Thank you all , Its a yes no question, But I were interested n how to prove it..
    Well I've tried this ... even before I posted the question, But I could not get to any solution.
    Now I've tried again and made it
    then , x+z = z+y = y+z = 0
    Thank you.

    Thank you, I've also tried this before I've posted the question.. But for some reason I didn't made it..
    After you told me you did it I looked at it again and it work..
    Thank you
    (R3-R2) -> (R3+R1) -> (R1-0.5R3) -> (R2-R1) Done...
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