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Problem: If $Ax = b$ has more than one solution so does $Ax = 0$, where $A$ is $m\times n$ real matrix.

In the explanation part it is written that when $Ax = b$ is consistent the solution sets of non-homogeneous equation and the homogeneous equation are translates of each other. So, in this case, the two equations have the same number of solutions.

I am not able to understand what exactly above explanation says?

However, I was thinking that if $Ax = b$ has more than one solution then it must be infinite. Hence there must be at least one free variable. Thus corresponding homogeneous system must be having infinite solutions. Am I thinking correctly?

Thanks for the help.

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  • $\begingroup$ The point is that, if $Ax=0$ has a solution, since the solution set is a linear subspace, it should contain the line spanned by that solution and so infinite if the ground field is infinite. $\endgroup$
    – Yai0Phah
    Jun 27, 2014 at 12:25
  • $\begingroup$ I didn't read your question. Consider two distinct solutions $u,v$ of $Ax=b$. it holds that $A(u-v)=Au-Av=b-b$. This gives you a non-null solution to $Ax=\bf 0$. $\endgroup$
    – Git Gud
    Jun 27, 2014 at 12:25
  • $\begingroup$ @GitGud Thank you I understood now. I also want to know whether my explanation is correct? $\endgroup$
    – Srijan
    Jun 27, 2014 at 12:28
  • $\begingroup$ @FrankScience Thank you very much. $\endgroup$
    – Srijan
    Jun 27, 2014 at 12:28

2 Answers 2

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There are too many words here, this is quite simple:

if $Ax = b = Ax_1$, then $A\times 0 = 0 = A(x-x_1)$. If $x\neq x_1$ then there are (at least) two different solutions to $Ax = 0$: $0$ and $x-x_1$ (and by the way, for any scalar $\lambda$, $\lambda(x-x_1)$ is also a solution).

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  • $\begingroup$ Thanks for the answer. Could you please tell me whether my explanation is correct or not? $\endgroup$
    – Srijan
    Jun 27, 2014 at 12:30
  • $\begingroup$ your explanation seems correct but not that precise (blabla is less precise than simple equation ;)). $\endgroup$
    – mookid
    Jun 27, 2014 at 12:32
  • $\begingroup$ :) Got it. Thanks a ton. $\endgroup$
    – Srijan
    Jun 27, 2014 at 12:32
  • $\begingroup$ thnaks :) {} {} {} $\endgroup$
    – mookid
    Jun 27, 2014 at 12:42
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Let $x$ and $y$ be two solutions of $Ax = b$, with $x\not= y$. Then: $$A(x-y) =Ax - Ay = b - b = 0$$so $x-y$ is a non-trivial solution of $Ax = 0$, and $\lambda(x-y)$ is a solution for every $\lambda \in \mathbb{R}$ so there are infinitely many.

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