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February 9th, 2012, 03:15 AM   #1
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Ax=B, clarification

Hey guys, this is a part of a proof that bothers me.

Assume AX=B has more then one solution, and let X(subscript 0)= X(subscript 1) - X(subscript 2). Where X(subscript 1) and X(subscript 2) are any two distinct solutions. Because X(subscript 1) and X(subscript 2) are distinct, the matrix X(subscript 0) is nonzero:

Ax(subscript 0) = A(X(subscript 1) - X(subscript 2)) = AX(subscript 1) - AX(subscript 2) = B-B = 0.

....

I get everything except this part. Why are we allowed to assume that a non-zero X(subscript 0) exists that makes AX(subscript 0) = 0? Also why can we assume that this non zero X(subscript 0) = X(subscript 1) - X(subscript 2), two other distinct solutions?

Thanks!
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February 11th, 2012, 03:19 PM   #2
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Re: Ax=B, clarification

Quote:
Originally Posted by Student444
. . .

Assume AX=B has more then one solution, . . .
That doesn't make sense to me.

If an [A](x) = (b) system has a solution, then it is unique.
Otherwise, it doesn't have a solution. It is underdetermined or overdetermined. In which case, it would be a problem for a Linear Least-Squares best fit, or some other approach.
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February 12th, 2012, 04:05 AM   #3
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Re: Ax=B, clarification

Quote:
Originally Posted by TomF
If an [A](x) = (b) system has a solution, then it is unique.
Otherwise, it doesn't have a solution.
For the number of solutions of a linear system in the form there exists three possibilities:
1. the system has no solution,
2. the system has a unique solution,
3. the system has infinitely many solutions.



Quote:
Originally Posted by Student444
...........Assume AX=B has more then one solution ............
This means that there exist such that: .

We denote their deference as i.e .

So we have i.e is a solution of .

Since we have i.e is a non-zero solution of .
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