- #1
Mesud1
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Suppose that I have an overdetermined equation system in matrix form:
Ax = b
Where x and b are column vectors, and A has the same number of rows as b, and x has less rows than both.
The least-squares method could be used here to obtain the best possible approximative solution. Let's call this solution "c".
Now, suppose I multiply some row of the equation system with a constant k. Let's say this row is the second row. In that case, I must multiply the 2nd row of A with k, as well as the 2nd row of b. This yields a new equation system, let's write it as:
Bx = d
If I use the method of least squares on the second system, I get a new solution that is different from c. Why is the solution different? Since I performed an elementary row operation on the first system to obtain the second system, shouldn't the two systems be equivalent, and therefore have the same least-squares solution?
When I did the same thing with a consistent system, I got the same solution for both systems.
Ax = b
Where x and b are column vectors, and A has the same number of rows as b, and x has less rows than both.
The least-squares method could be used here to obtain the best possible approximative solution. Let's call this solution "c".
Now, suppose I multiply some row of the equation system with a constant k. Let's say this row is the second row. In that case, I must multiply the 2nd row of A with k, as well as the 2nd row of b. This yields a new equation system, let's write it as:
Bx = d
If I use the method of least squares on the second system, I get a new solution that is different from c. Why is the solution different? Since I performed an elementary row operation on the first system to obtain the second system, shouldn't the two systems be equivalent, and therefore have the same least-squares solution?
When I did the same thing with a consistent system, I got the same solution for both systems.