TITLE: Prove that if $b \ne 0$ then the set of solutions to $Ax=b$ is not a subspace
QUESTION [8 upvotes]: I realize that I will probably have to prove that the solution set does not contain the zero vector.  I've been trying to prove this, but I am not sure how to.  
This is what I have so far, but it doesn't sound very proofy.  I'm new to proofs and math, so I don't know if this is right or not.
My attempt?
Assume that $A$ is a non-zero invertible matrix.  Then $0 = A^{-1}b$ is impossible unless $b=0$.  Therefore, the set is not a subspace.

REPLY [6 votes]: Your assumption that $A$ is invertible is unnecessary. Instead, consider proving the contrapositive of the given implication:

If $S = \{\vec x \in \mathbb R^n \mid A\vec x = \vec b\}$ is a subspace, then $\vec b = \vec 0$.

Indeed, since $S$ is a subspace, we know that $\vec 0 \in S$. But then $\vec b = A\vec 0 = \vec 0$, as desired. $~~\blacksquare$

REPLY [2 votes]: Let $x={\bf 0}$. $A(x+x)=Ax+Ax=b+b\neq b$ if $b\neq{\bf 0}$.

REPLY [2 votes]: You are right: try and prove that the subspace does not contain the zero vector. If $x$ is the zero vector then $Ax$ is a matrix full of zeroes. But $b\ne 0$...