Prove that $A$ is affine iff, for every $a\in A$, $A-\{a\}$ is a subspace of $\mathbb{R}^n$

I know that to show a subspace I must show $\forall x,y\in A-\{a\}, x+y\in A-\{a\}$ and $\forall \alpha\in\mathbb{R}, x\in A-\{a\}, \alpha x\in A-\{a\}$.

And I believe I can work out how to do that part, but I am struggling with the other direction. We know $A-\{a\}$ is a subspace and need to prove $A$ is affine.

  • $\begingroup$ Subspaces should always contain $0$, so I think you need a different phrase here. $\endgroup$ – user42761 Mar 2 '17 at 19:48
  • $\begingroup$ To prove the other direction, I would use the caracterization by barycenters. $\endgroup$ – nicomezi Mar 2 '17 at 19:55
  • 1
    $\begingroup$ The title is irritating, as $-$ is the element wise vector subtraction here and not set difference. $\endgroup$ – user251257 Mar 2 '17 at 20:01

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