February 9th, 2012, 03:15 AM  #1 
Newbie Joined: Feb 2012 Posts: 19 Thanks: 0  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) = BB = 0. .... I get everything except this part. Why are we allowed to assume that a nonzero 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! 
February 11th, 2012, 03:19 PM  #2  
Newbie Joined: Jan 2008 Posts: 22 Thanks: 0  Re: Ax=B, clarification Quote:
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 LeastSquares best fit, or some other approach.  
February 12th, 2012, 04:05 AM  #3  
Newbie Joined: Mar 2011 Posts: 10 Thanks: 0  Re: Ax=B, clarification Quote:
1. the system has no solution, 2. the system has a unique solution, 3. the system has infinitely many solutions. Quote:
We denote their deference as i.e . So we have i.e is a solution of . Since we have i.e is a nonzero solution of .  

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