What is the significance of the proof for matrix solutions?

In summary, the theorem states that a system of linear equations has either zero, one, or infinitely many solutions. If Ax=b has two solutions, then Ax0=b and Ax1=b.
  • #1
icesalmon
270
13

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
 
Physics news on Phys.org
  • #2
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?
 
  • #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.
 
  • #4
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:
  • #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.
 
  • #6
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:
  • #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?
 
  • #8
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.
 

FAQ: What is the significance of the proof for matrix solutions?

What is a matrix solution?

A matrix solution is a set of values that satisfy a system of linear equations represented by a matrix. It is a set of numbers that, when substituted into the equations, make all the equations true.

How do you find a matrix solution?

To find a matrix solution, you can use methods such as Gaussian elimination or Cramer's rule. These methods involve manipulating the matrix and its coefficients to solve for the variables and obtain the solution.

Can a matrix have more than one solution?

Yes, a matrix can have more than one solution. This is known as an infinite number of solutions or a dependent system, where the equations are not enough to uniquely determine the values for the variables.

How do you know if a matrix has no solution?

If a matrix has no solution, it is known as an inconsistent system. This means that the equations are contradictory and cannot be satisfied by any set of values for the variables. This can be determined by performing row operations on the matrix and reaching a row of all zeros.

What is the importance of matrix solutions in science?

Matrix solutions are important in science because they can be used to solve systems of equations that arise in many fields, such as physics, engineering, and economics. They also have applications in computer science, data analysis, and other areas. Matrix solutions allow for the modeling and analysis of complex systems and can provide insights and solutions to real-world problems.

Similar threads

Linear Algebra uniqueness of solution
Replies
6
Views
1K
How to find a system of equations when the solution is given?
Replies
6
Views
380
Efficient Prediction of Linear Congruential Pseudorandom Numbers
Replies
1
Views
1K
[Linear Algebra] Closed formula for recursive sequence
Replies
8
Views
2K
Solution Sets of Linear Systems
Replies
14
Views
2K
Matrix Relative to B and B' R3 to R3
Replies
1
Views
1K
Find the basis of a vector subspace of R^2,2
Replies
9
Views
2K
Linear Algebra: Non-Singular Matrix and Zero Matrix
Replies
13
Views
2K
Proving Inequality in Inner Product Spaces
Replies
4
Views
2K
Proof of a system of linear equations
Replies
4
Views
3K

Hot Threads

Recent Insights

Back
Top