**Edited by kevop, 26 December 2012 - 08:25 AM.**

**1**

# Changing roles in linear system for stability

###
#1
Members - Reputation: **111**

Posted 26 December 2012 - 06:37 AM

###
#3
Members - Reputation: **111**

Posted 26 December 2012 - 12:20 PM

I want to test the stability of system using a non-invertible matrix B of the system instead of A, when X is unknown and A, B are both known.

I am not interested in the solution directly but want to reduce error in the resulting solution by reducing condition number of B instead of A.

All of them are matrices: X = 4 x M, A = 4 x N, B = N x M.

###
#5
Members - Reputation: **111**

Posted 26 December 2012 - 12:51 PM

All of them are matrices: X = 4 x M, A = 4 x N, B = N x M.That doesn't seem right. If you can multiply A times X, the number of columns in A has to match the number of rows in X.

The transpose can take care of such issues. They are not important but if you wish it can be rewritten as:

A = N x 4

X = 4 x M

B = N x M

###
#6
Members - Reputation: **1820**

Posted 26 December 2012 - 04:21 PM

There are an infinite number of solutions to what you are trying to solve when N < 4. For example, let N and M be 1. That leaves us with 1x4 * 4x1 = 1x1. So Ax=b would basically be b = A_{1} * x_{1} + A_{2} * x_{2} + A_{3} * x_{3} + A_{4} * x_{4}. For any given value of b, there are an infinite number of solutions of x that satisfy Ax=b. This is the case for any value of N < 4. When N = 4, x can be solved simply by calculating A^{-1}b. When N > 4 and M=1, you have an overdetermined system. You can use Least Squares for that. I am not sure if least squares will work for you if M > 1.

EDIT: Upon further thought, I think least squares is what you are looking for. It should work when M > 1.

**Edited by HappyCoder, 26 December 2012 - 04:26 PM.**

###
#7
Members - Reputation: **111**

Posted 27 December 2012 - 12:46 AM

Well I am not interested in the solution which is trivial to obtain but in the quality of future solutions.

Let me put it this way:

There are a total of H locations at which observations can be made.

Then if I have initially N locations in A (N >=4) at a given time then which of location from set H that are not in A will be most optimal for addition to A at N+1.

That is when you use condition numbers to test the quality of possible future solutions for addition into A.

So the point is to test the quality of the future solution not using location in A but using the observation in B. By checking which one observation from H can improve the quality of future solution without ever solving the least square itself.

However checking condition number of B would be incorrect as it is not involved in inversion when finding X.

So there must a reason to relate condition number of A and B together for this linear system.