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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.

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  • $\begingroup$ 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? $\endgroup$ – Git Gud Oct 27 '14 at 23:48
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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. $~~\blacksquare$

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  • $\begingroup$ Very interesting. Much simpler than my attempt. But, would my attempt also be considered correct as well? $\endgroup$ – Jason Oct 27 '14 at 23:54
  • $\begingroup$ Your attempt is incomplete. What if $A$ is not invertible? $\endgroup$ – Adriano Oct 27 '14 at 23:56
  • $\begingroup$ 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. $\endgroup$ – Jason Oct 28 '14 at 0:07
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Let $x={\bf 0}$. $A(x+x)=Ax+Ax=b+b\neq b$ if $b\neq{\bf 0}$.

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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 $b\ne 0$...

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