Affine space/Affine subspace/Characterization/Fact/Proof

Proof

If ${}F$ is empty, then all three conditions are true. So we assume that ${}F$ is not empty. ${}(1)\Rightarrow (2)$. Let ${}F=P+U$ with ${}P\in F$ and a linear subspace ${}U\subseteq V$. Then, ${}P_{i}=P+u_{i}$ with some ${}u_{i}\in U$. Due to the definition of a barycentric combination, it follows that

${}\sum _{i=1}^{n}a_{i}P_{i}=P+\sum _{i=1}^{n}a_{i}{\overrightarrow {PP_{i}}}=P+\sum _{i=1}^{n}a_{i}u_{i}\,$

is an element of ${}F$.

${}(2)\Rightarrow (3)$. This is a weakening of the condition.

${}(3)\Rightarrow (1)$. We choose a point ${}P\in F$, and consider

${}U:={\left\{{\overrightarrow {PQ}}\mid Q\in F\right\}}\subseteq V\,.$

We have ${}0\in U$. For ${}Q,Q'\in F$, due to the condition, also ${}-P+2Q$ and ${}-P+2Q'$ belong to ${}F$. Therefore, also

${}{\frac {1}{2}}{\left(-P+2Q\right)}+{\frac {1}{2}}{\left(-P+2Q'\right)}=-P+Q+Q'\,$

belongs to ${}F$, where this equality rests on exercise. This point equals

${}P-{\overrightarrow {PP}}+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}=P+{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}\,,$

so that ${}{\overrightarrow {PQ}}+{\overrightarrow {PQ'}}$ belongs to ${}U$. Hence, ${}U$ is closed under the vector addition. Let ${}Q\in F$ and ${}s\in K$. Then, due to the condition, also

${}(1-s)P+sQ=P+s{\overrightarrow {PQ}}\,$

belongs to ${}F$, and, therefore, ${}s{\overrightarrow {PQ}}$ belongs to ${}U$. Thus, ${}F=P+U$ with a linear subspace ${}U$.

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