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Real Affine Sets
Real Subspaces
Real Translates
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Real Affine Sets and Subspaces


What are the affine sets in terms of subspaces?

Affine sets which are subspaces

The subspaces of $\R ^n$ are the affine sets which contain the origin.

$M \subset \R ^n$ is a subspace if and only if $M$ is affine and $0 \in M$.
($\Rightarrow$) Suppose $M$ is a subspace. Then $0 \in M$. Also $\alpha x + \beta y \in M$ for all $\alpha , \beta \in \R $ and $x, y \in \R ^n$ In particular, $(1- \lambda )x + \lambda y \in M$ for all $\lambda \in \R $, $x, y \in \R ^n$. In other words, $M$ contains the line through $x$ and $y$. ($\Leftarrow$) Suppose $M$ is affine and $0 \in M$. $M$ is closed under scalar multiplication since

\[ \alpha x = (1 - \alpha )0 + \alpha x \]

is in the line through $0$ and $x$.
$M$ is closed under vector addition since

\[ (1/2)(x + y) = (1-1/2)x + (1/2)y \]

is in the line through $x$ and $y$.
Thus, $x+y = 2(1/2)(x+y) \in M$.

Affine sets as translated subspaces

Suppose $M \neq \varnothing$ is affine. Then there exists a unique subspace $L$ and vector $a \in \R ^n$ for which $M = L + a$. Moreover,

\[ L = M - M = \Set{x - y}{x,y \in M}. \]

First, uniqueness. Suppose $L_1$ and $L_2$ are subspaces parallel to $M$. We will show that $L_2 \supset L_1$ (and similarly, $L_1 \supset L_2$). Since $L_1$ and $L_2$ are both parallel to $M$, they are also parallel to each other. Consequently, there exists $a \in \R ^n$ with $L_2 = L_1 + a$. Since $0 \in L_2$ (it is a subspace, after all), $-a \in L_1$. Since $L_1$ is a subspace, $a \in L_1$. So $x + a \in L_1$ for every $a \in L_1$, and so $L_2 = L_1 + a \subset L_1$. A similar argument gives $L_1 \supset L_2$. If $y \in M$, then $M + (-y) = M - y$ is a translate of $M$ containing zero (since $y - y = 0$). In other words, the affine set $M - y$ is a subspace. This, then, is the unique subspace parallel to $M$. Since $y$ was arbitrary, the subspace parallel to $M$ is $L = \cup_{y \in M} M - y = \Set{x - y}{x,y \in M}$.
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