What is the significance of the proof for matrix solutions?

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Homework Help Overview

The discussion revolves around understanding the proof of Theorem 1.6.1, which states that a system of linear equations can have either zero, one, or infinitely many solutions. Participants are exploring the implications of the proof, particularly focusing on the case where there are infinitely many solutions.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants are attempting to grasp the significance of the proof, particularly the reasoning behind the manipulation of solutions and the role of scalar multiplication in generating new solutions. Questions arise regarding the necessity of certain steps in the proof and the implications of having multiple solutions.

Discussion Status

The discussion is ongoing, with participants expressing confusion about specific aspects of the proof and its significance. Some have begun to recognize that having two solutions implies the existence of infinitely many solutions, while others are questioning the mathematical reasoning behind generating new solutions from existing ones.

Contextual Notes

Participants are grappling with the abstract nature of the proof and its implications, indicating a need for deeper exploration of the concepts involved. There is a mention of potential geometric interpretations, suggesting that the discussion may evolve towards visualizing the solutions in a geometric context.

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Homework Statement


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
 
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icesalmon said:

Homework Statement


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

Try thinking of this way. If there are two solutions Ax0=b and Ax1=b is (x0+x1)/2 a solution?
 
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.
 
icesalmon said:
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.

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:
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.
 
icesalmon said:
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.

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:
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?
 
icesalmon said:
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?

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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