# Prove that if $b \ne 0$ then the set of solutions to $Ax=b$ is not a subspace

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.

• You're being asked to prove that $\{x\colon Ax=b\}$ is not a vector space and you correctly guessed that $\bf 0$ isn't in this set. How to prove it? Assume it is, then replace $x$ by $\bf 0$, what do you get? – Git Gud Oct 27 '14 at 23:48

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. $~~\blacksquare$

• Very interesting. Much simpler than my attempt. But, would my attempt also be considered correct as well? – Jason Oct 27 '14 at 23:54
• Your attempt is incomplete. What if $A$ is not invertible? – Adriano Oct 27 '14 at 23:56
• True, then my attempt wouldn't cover that. I guess my proof doesn't prepare for the possibility that there exists a non-invertible matrix that doesn't satisfy the question. – Jason Oct 28 '14 at 0:07

Let $x={\bf 0}$. $A(x+x)=Ax+Ax=b+b\neq b$ if $b\neq{\bf 0}$.

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 $b\ne 0$...