In $Ax = b$, why does $x$ not pass through the origin? I am facing some difficulty in understanding the concept of $x$ being a vector which doesn't pass through the origin.
Let me tell you what I understand:

*

*A subspace is a space that contains all linear combinations of a vector and it passes through the origin.


*$Ax = b$, $A$ is invertible and $b$ is the set of solutions for $x$ vector of unknowns.
In Prof. Gilbert Strang's book, there is a mention of the following. Any vector $x_n$ in the null space can be added to a particular solution $x_p$. The solutions to all linear equations have this form, $x = x_p +x_n$:
Complete solution $Ax_p = b$ and $Ax_n = 0$ produce $A(x_p +x_n) = b$.
So is this, complete solution, a plane? It certainly is a vector space because all linear combinations of A result in b and if so, then it should pass through the origin.
If my understanding of a vector space is correct, then why does $x$ not pass through the origin?
Please feel free to correct me wherever you feel necessary, I am unable to understand this at all. It is quite impossible for $x$ to be 0 and $b$ to be non-zero at the same time yet the idea of the vector space, origin and null space is a bit confusing for me.
 A: 
It certainly is a vector space

No, it isn't. Let $x_1$ and $x_2$ be two solutions of $Ax=b$. Then, $A(x_1+x_2)=Ax_1+Ax_2 = b+b=2b \neq b$.
In case $b=0$, the set of solutions to $Ax=0$ is indeed a vector subspace (the nullspace of $A$).
A: Claim: Suppose the system $Ax = b$ has a particular solution $x_p$. Then the set $S$ of all solutions to $Ax = b$ equals $x_p + N(A)$, where $N(A)$ is the nullspace of $A$. (Here the notation $x_p + N(A)$ means the set $\{ x_p + n: n \in N(A) \}$.)
Proof: To prove that the sets are equal, we show containment in both directions.
$S \subseteq x_p + N(A)$: Let $s$ be an arbitrary element of $S$; in other words, $As = b$. Since $Ax_p = b$ as well, we have $As = Ax_p$. Hence $As - Ax_p = 0$, i.e. $A(s - x_p) = 0$. So $s - x_p \in N(A)$, which means that $s - x_p = n$ for some $n \in N(A)$. So $s = x_p + n \in x_p \in N(A)$, as desired.
$x_p + N(A) \subseteq S$: Let $x_p + n$ (where $n \in N(A)$) be an arbitrary element of $x_p + N(A)$. Then $A(x_p + n) = Ax_p + An = b + 0 = b$, so that $x_p + n$ solves the system $Ax = b$; in other words, $x_p + n \in S$, as desired. $\square$
Geometrically, the set of solutions to the consistent system $Ax = b$ is parallel to the nullspace (which is the set of solutions to $Ax = 0$). This is what $x_p + N(A)$ means geometrically: take all the vectors in the nullspace and translate them by the vector $x_p$. So, for example, if the nullspace is a plane in 3D, then the set of solutions to the consistent system $Ax = b$ is another plane parallel to it.
