Question about a null space theorem and its converse. The theorem states (quoted from my book)
"If $x_0$ is any solution of a consistent linear system Ax=b, and if S = {$v_1, v_2, ..., v_k$} is a basis for the null space of A, then every solution of Ax=b can be expressed in the form
$x = x_0 + c_1v_1 + c_2v_2+...+c_kv_k$
Conversely, for all choices of scalars $c_1, c_2, ..., c_k$, the vector x in this formula is a solution of Ax=b."
My question is regarding the converse. Does it imply that for any scalar $c_1, c_2, ..., c_k$ linear combination of Ax=b's null space, there will exist some solution $x_0$ for which $x$ will be a solution of Ax=b (of which $x_0$ is a solution as well)?
My impression from the wording in my book is that: if you have a solution $x_0$ from a consistent linear system Ax=b, then any linear combination of its null space plus $x_0$ will also produce a solution in the same system for which $x_0$ is consistent. I don't believe this interpretation is correct.     
 A: It is saying that for any choice of scalars, and any specific solution to the equation $Ax_0=b$ we have $Ax=b$. In other words, $x$ is a solution to the equation as well, for any choice of scalars.
This is equivalent to saying $Ax=b$ iff $Ax_0=b$. The proof of which can be done by letting $c=c_1v_1+...+c_nv_n$ be vector of size $x_0$. Then $A(x)=A(x_0+c)=Ax_0+Ac=Ax_0+0=Ax_0=b$.
A: Your interpretation is, I think, correct, though I can't understand what does "...for which $\;x_o\;$ is consistent" can possibly mean within this context.
This is one of the facts that, at least in my case, made get in love even more with mathematics while in university: if we have a non-homogeneous system $\;Ax=b\;$ , for which we know a particular solution $\;x_0\;$ , meaning: $\;Ax_0=b\;$, and if $\;P:=\{\;x\;:\;\;Ax=0\;\}\;$ is the solution subspace of the corresponding homogenoeus system, then the solution set (i.e., all the solutions) of the non-homogeneous original system is $\;x_0+P\;$
A: Your impression is indeed correct.
If $x_0$ is a solution, then $Ax_0 = b$.
As $S$ is a basis of the null space of $A$, any linear combination of the vectors of $S$ is also in the null space of $A$, so $A(c_1x_1 + \cdots + c_k x_k) = 0$.
Then, for $x = x_0 + c_1x_1 + \cdots c_k x_k$, we have:
$Ax = Ax_0 + 0 = b$,
meaning $x$ is also a solution.
