Let $A$ be an $m \times n$ matrix and let $x, b$ be $n \times 1$ vectors. Is it true that, if the equation $Ax = b$ has more than one solution, then $Ax = 0$ has more than one solution as well?
I was thinking of an argument along the lines of: if $Ax=b$ has many solutions, then the reduced row echelon form of $A$ has at least one non-pivot entry (associated to a free variable of the system). So $Ax = 0$ will also be equivalent to a system with a free variable, therefore having more than one solution.
Is this reasoning correct?
Suppose $x_h$ represents all solutions to the equation $Ax=0$, namely, $Ax_h=0$.
and $x_p$ is a particular solution to the equation $Ax=b$, namely, $Ax_p=b$,
then $x_h+x_p$ is also a solution to this equation, because
$$A(x_h+x_p)=Ax_h+Ax_p=0+b=b$$
Since the dimension of the null space is $\ge1$, which means there are infintely many solutions for $x_h$, hence infinitely many solutions for $x_h+x_p$