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Linear Algebra Nonparallel vector Proof

  1. Aug 29, 2011 #1
    1. The problem statement, all variables and given/known data
    Suppose x, y [itex]\in\Re^{n}[/itex] are nonparallel vectors.
    a) Prove that if sx+ty=0, then s=t=0
    b) Prove that if ax+by=cx+dy, then a=c and b=d


    2. Relevant equations



    3. The attempt at a solution
    I'm very new to proof-based math, so I'm just trying to get my feet wet here. I realize this is a very simple problem, but I was hoping someone could walk me through it.

    Assuming s[itex]\neq[/itex]0, then sx=-ty. Thus, x=(-t/s)y, which goes against the definition of nonparallel vectors. (Two vectors are parallel if they can be written with scalar constant. i.e. x=cy or y=cx). So, as a result, we know s[itex]\neq[/itex]0 must not be possible. Using similar logic, t=0 cannot be possible, either. Therefore, s=t=0.

    I realize this is incredibly rough; I'm just looking to see if my thinking is on the right track.

    Thanks!
     
  2. jcsd
  3. Aug 29, 2011 #2

    Mark44

    Staff: Mentor

    You're on the right track. What you're doing is a proof by contradiction, which is where you prove that P == Q by assuming that P is true and that Q is false. If you arrive at a contradiction, then you have indirectly proved that P ==> Q.

    For this problem you should assume that both s and t are nonzero. That way you can show that x = (-t/s) y and that y = (-s/t) x. Since, by assumption x and y are nonparallel, you have arrived at a contradiction.
     
  4. Aug 29, 2011 #3

    Mark44

    Staff: Mentor

    For the second part, group the x terms together and the y terms together.
     
  5. Aug 29, 2011 #4
    Notice also that for the second problem, you may find it helpful to use the result of the previous problem. The proof will follow quite naturally from this. Good work so far!
     
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