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Proof about matrix solutions

  1. Jan 23, 2013 #1
    1. The problem statement, all variables and given/known data
    I do not need to know if this is correct, I am trying to understand the proof.

    Theorem 1.6.1: A system of equations has either zero, one, or infinitely many solutions. There are no other possibilities.

    If Ax = b is a system of linear equations, exactly one of the following is true (a) the system has no solutions, (b) the system has exactly one solution, or (c) the system has more than one solution. The proof will be complete if we can show that the system has infinitely many solutions in case (c).

    i). Assume Ax = b has more than one solution, and let x0 = x1 - x2, where x1 and x2 are any two distinct solutions. Because x1 and x2 are distinct, the matrix x0 is nonzero, moreover: Ax0 = A(x1-x2) = b - b = 0

    ii). If we let k be any scalar, then A(x1 + kx0) = b + kx0 = b which shows that x1 + kx0 is a solution of Ax = b. Since x0 is nonzero and there are infinitely many choices for k, the system Ax = b has infinitely many solutions.

    I'm not really seeing the significance of part 1. I find ONE solution for b, 0. As for the second part, I feel like I should only be concerned with this as a proof to Ax = b having infinitely many solutions. If we just multiply x1 by k and just change the k value I will change what kx1 is. Why do I attach k onto x0 if it's just going to sum to zero later on, i'm not getting a different answer from x0 by multiplying by any number.
    Sorry if this doesn't make much, if any, sense. Thanks anyway for the help
     
  2. jcsd
  3. Jan 23, 2013 #2

    Dick

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    Try thinking of this way. If there are two solutions Ax0=b and Ax1=b is (x0+x1)/2 a solution?
     
  4. Jan 23, 2013 #3
    i'm not sure, but if x0 = x1-x2 then x0 + x1 = 2x1 - x2
    dividing that by 2 doesn't seem to do anything significant to the problem so I will say yes.
     
  5. Jan 23, 2013 #4

    Dick

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    Well, sure if Ax0=b and Ax1=b then A((x0+x1)/2)=Ax0/2+Ax1/2=b/2+b/2=b. So (x0+x1)/2 is another solution. That means if you have two solutions then you must have three solutions, x0, x1 and (x0+x1)/2. See where this is going?
     
    Last edited: Jan 23, 2013
  6. Jan 23, 2013 #5
    not at all, I mean I understand arithmetically that (A( xn + xn+1 )) / 2 = b, so b= b. but I could just say (A( 2xn + xn+1 )) / 3 would work as well. This feels like i'm just making up some mathematical form to suit the problem, it doesn't feel like i'm understanding what you're trying to tell me.
     
  7. Jan 23, 2013 #6

    Dick

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    The goal of the problem is to show that if you have more than one solution then you have an infinite number of solutions, isn't it? Having zero solutions is possible. Having one solution is possible. Having exactly two solutions is not possible. Nor is three or four etc. The only other possibility is an infinite number.
     
    Last edited: Jan 23, 2013
  8. Jan 24, 2013 #7
    I understand, so once you have two solutions you can generate an infinite number of solution producing expressions using them.

    Is there a geometrical point of view to what you've been explaining?
     
  9. Jan 24, 2013 #8

    Dick

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    Sure, in the proof they talk about the set x1+k*(x2-x1). As you vary k that point moves along the line through x1 and x2.
     
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