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

linear algebramatrices

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.

Best Answer

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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