My Math Forum Ax=B, clarification

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 February 9th, 2012, 04:15 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) = 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!
February 11th, 2012, 04:19 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.

February 12th, 2012, 05:05 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 $AX=B$ 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 $X_1 \neq X_2$ such that: $AX_1= B, \, AX_2=B$.

We denote their deference as $X_0$ i.e $X_1 -X_2= X_0$.

So we have $AX_0= A(X_1 - X_2) = AX_1 - AX_2 = B-B = O$ i.e $X_0$ is a solution of $AX=O$.

Since $X_1 \neq X_2$ we have $X_0 \neq O$ i.e $X_0$ is a non-zero solution of $AX=O$.

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