Characterization of Affine Transformations

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Theorem

Let $\EE$ and $\FF$ be affine spaces over a field $K$.

Let $\LL: \EE \to \FF$ be a mapping.


Then $\LL$ is an affine transformation if and only if:

$\forall p, q \in \EE: \forall \lambda \in K: \map \LL {\lambda p + \paren {1 - \lambda} q} = \lambda \map \LL p + \paren {1 - \lambda} \map \LL q$

where $\lambda p + \paren {1 - \lambda} q$ and $\lambda \map \LL p + \paren {1 - \lambda} \map \LL q$ denote barycenters.


Proof

Sufficient Condition

Let $\LL$ be an affine transformation.

Let $L$ be the tangent map.

Let $r \in \EE$ be any point.

Then by definition we have:

$\lambda p + \paren {1 - \lambda} q = r + \lambda \vec{r p} + \paren {1 - \lambda} \vec{r q}$

Thus we find:

\(\ds \map \LL {\lambda p + \paren {1 - \lambda} q}\) \(=\) \(\ds \map \LL r + \map L {\lambda p + \paren {1 - \lambda} q}\) Definition of Affine Transformation
\(\ds \) \(=\) \(\ds \map \LL r + \lambda \map L p + \paren {1 - \lambda} \map L q\) since $L$ is linear
\(\ds \) \(=\) \(\ds \lambda \map \LL p + \paren {1 - \lambda} \map \LL q\) Definition of Barycenter

$\Box$


Necessary Condition

Suppose that for all points $p, q \in \EE$ and all $\lambda \in \R$:

$\map \LL {\lambda p + \paren {1 - \lambda} q} = \lambda \map \LL p + \paren {1 - \lambda} \map \LL q$

Let $E$ be the difference space of $\EE$.

Fix a point $p \in \EE$, and define for all $u \in E$:

$\map L u = \map \LL {p + u} - \map \LL p$

Let $q = p + u$.

Then:

$\map \LL q = \map \LL p + \map L u$

So to show that $\LL$ is affine, we are required to prove that $L$ is linear.

That is, we want to show that for all $\lambda \in k$ and all $u, v \in E$:

$\map L {\lambda u} = \lambda \map L u$

and:

$\map L {u + v} = \map L u + \map L v$

First of all:

\(\ds \map L {\lambda u}\) \(=\) \(\ds \map \LL {p + \lambda u} - \map \LL p\) Definition of $L$
\(\ds \) \(=\) \(\ds \map \LL {\paren {1 - \lambda} + \lambda \paren {p + u} }\) Definition of Barycenter
\(\ds \) \(=\) \(\ds \paren {1 - \lambda} \map \LL p + \lambda \map \LL {p + u} - \map \LL p\) by hypothesis on $\LL$
\(\ds \) \(=\) \(\ds \lambda \paren {\map \LL {p + u} - \map \LL p}\)
\(\ds \) \(=\) \(\ds \lambda \map L u\) Definition of $L$

Now it is to be shown that

$\map L {u + v} = \map L u + \map L v$

First:

$p + u + v = \dfrac 1 2 \paren {p + 2 u} + \dfrac 1 2 \paren {p + 2 v}$

Now:

\(\ds \map \LL {p + u + v}\) \(=\) \(\ds \map \LL {\frac 1 2 \paren {p + 2 u} + \frac 1 2 \paren {p + 2 v} }\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map \LL {p + 2 u} + \frac 1 2 \map \LL {p + 2 v}\) by hypothesis on $\LL$
\(\ds \) \(=\) \(\ds \frac 1 2 \paren {\map \LL {p + 2 u} - \map \LL p} + \frac 1 2 \paren {\map \LL {p + 2 v} - \map \LL p} + \map \LL p\)
\(\ds \) \(=\) \(\ds \frac 1 2 \map L {2 u} + \frac 1 2 \map L {2 v} + \map \LL p\) Definition of $L$
\(\ds \) \(=\) \(\ds \map L u + \map L v + \map \LL p\) as $L$ preserves scalar multiples

From the above calculation:

$\map L {u + v} = \map \LL {p + u + v} - \map \LL p = \map L u + \map L v$

This shows that $L$ is linear, and therefore concludes the proof.

$\blacksquare$

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