# Affine subspace iff statement

I cannot find a proof for the following statement, though the statement itself seems common enough:

$X$ is an affine subspace of $\mathbb{R}^n \iff \forall a\in X: X - \{a\}$ is a linear subspace.

An affine subspace being a subset of a vector space, such that any affine combination $\sum_{i=1}^k \lambda_i x^i,\ \sum_{i=1}^k \lambda_i = 1$ of vectors from $X$ also belongs to $X$.

Could someone supply a proof?

• What's your definition of affine subspace, is it a subset of $\mathbb{R}^n$ that's closed under affine combinations? – littleO Oct 31 '12 at 22:14
• Usually $X-\{a\}$ is the set $X$ without the point $a$, and I assume you want the translate of $X$ by $-a$. This is sometimes denoted by $X-a$, but for clarity I would explain what you mean, i.e., that it is the set of all $x-a$ where $x\in X$. – Lukas Geyer Oct 31 '12 at 22:34

Let $X$ be an affine subspace and $a \in X$. We will prove that $X - a$ is a linear subspace, so let $x,y \in X$ and $\lambda, \mu \in \mathbb R$. Then $\lambda(x-a) + \mu(y-a) = \lambda x + \mu y + (1 -\lambda - \mu)a - a$ Now $\lambda x +\mu y + (1-\lambda-\mu)a \in X$ as an affine combination of vectors from $X$, so $\lambda(x-a) + \mu(y-a) = \lambda x + \mu y + (1 -\lambda - \mu)a - a \in X - a$ and $X-a$ is closed under linear combinations, hence a linear subspace.
Now suppose that $X-a$ is a linear subspace for some $a \in X$, then for $x_i \in X$ and $\lambda_i \in \mathbb R$ with $\sum_i \lambda_i = 1$ we have \begin{align*} \sum_i \lambda_i x_i &= \sum_i\lambda_i (x_i - a) + \sum_i\lambda_i a\\ &= \sum_i \lambda_i (x_i - a) + a\\ &\in X - a + a\\ &= X. \end{align*} Hence, $X$ is affine.
Note, that in arguing, we always supposed $X \ne \emptyset$, if $X$ is the empty affine subspace, the right hand side is trivially true.