Finding more than one solution for Matrix Multiplication (Ax = b)

For some A and b, there are infinite solutions.

Currently, I only know one way to solve for x, which solves for only 1 x. That is, A\b. How do I find more than 1 solution? Is there any I can find specifically 2 solutions?

Thanks, Clark

Answer 1

[Wayne King]

For example:

 A = [1 2; 1 2];
 b = [3 3]';
 x = pinv(A)*b; % one solution
 A*x
 % the nullspace of A has dimension 1. So just add that vector
 % to the solution to get another solution
 y = x+null(A);
 A*y  % another

Yet another

   z = 2*null(A);
   w = x+z;
   A*w

[Clark]

Ah! Thanks! I was trying to do A\b + null(null(A)) which obviously wasn't working. So, instead, scalar * null(A). Thanks!

Answer 2

[Wayne King]

Yes, you can use null() to find a vector from the null space

Answer 3

[Clark]

I don't understand how to use null for this purpose. If I use null(A\b) for example, the answer is simply an empty 1x0 matrix. Could you please elaborate? Sorry, I'm a beginner. I don't see what null would have to do with this. I'm trying to figure out how to use it anyhow.

Thank you.

[Wayne King]

No, you just use null() on the matrix. null(A) then you get a basis for the nullspace of A, you can add any linear combination of vectors from the nullspace of A to a particular solution and you get a different solution of the linear system

Answer 4

[Clark]

Okay, so, A\b + null(A) gives me another solution.

But, can I find a third solution? I was expecting there may be some function which accepts an argument to return a certain number of solutions.

Thanks.

[Wayne King]

All the solutions will be a particular solution plus linear combinations of the nullspace. You can easily write a function that gives you a specified number of linear combinations.