Affine-linear mapping/Barycentric combination/Fact/Proof

Proof

Let ${}V$ and ${}W$ denote the vector spaces for ${}E$ and for ${}F$, respectively. Suppose first that ${}\psi$ is affine-linear with linear part

$\psi _{0}\colon V\longrightarrow W.$

Let a barycentric combination ${}\sum _{i\in I}a_{i}P_{i}$ with ${}P_{i}\in E$ and ${}\sum _{i\in I}a_{i}=1$ be given. Then we have (with an arbitrary point ${}Q\in E$)

${}{\begin{aligned}\psi {\left(\sum _{i\in I}a_{i}P_{i}\right)}&=\psi {\left(Q+\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\psi _{0}{\left(\sum _{i\in I}a_{i}{\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}\psi _{0}{\left({\overrightarrow {QP_{i}}}\right)}\\&=\psi (Q)+\sum _{i\in I}a_{i}{\overrightarrow {\psi (Q)\psi (P_{i})}}\\&=\sum _{i\in I}a_{i}\psi (P_{i}).\end{aligned}}$

Now, suppose that the mapping ${}\psi$ is compatible with barycentric combinations. We set

${}\varphi (v):={\overrightarrow {\psi (P)\psi (P+v)}}\,$

for ${}v\in V$, where ${}P\in E$ is any point. We first show that this is independent of the chosen point ${}P$. The sum

$(P+v)-(Q+v)+Q$

is a barycentric combination of the point ${}P$, see exercise. Therefore, we have in ${}F$ the equality

${}\psi (P+v)-\psi (Q+v)+\psi (Q)=\psi (P)\,.$

Hence, we have in ${}V$ the equality

${}{\overrightarrow {\psi (P)\psi (P+v)}}-{\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (P)\psi (Q)}}=0\,,$

and, therefore,

${}{\begin{aligned}{\overrightarrow {\psi (P)\psi (P+v)}}&={\overrightarrow {\psi (P)\psi (Q+v)}}-{\overrightarrow {\psi (P)\psi (Q)}}\\&={\overrightarrow {\psi (P)\psi (Q+v)}}+{\overrightarrow {\psi (Q)\psi (P)}}\\&={\overrightarrow {\psi (Q)\psi (Q+v)}}.\end{aligned}}$

We have to show that ${}\varphi$ is linear. For ${}u={\overrightarrow {PQ}}$ and ${}v={\overrightarrow {PR}}$, we have

${}{\begin{aligned}\psi (P)+\varphi (au+bv)&=\psi (P+au+bv)\\&=\psi {\left(P+a{\overrightarrow {PQ}}+b{\overrightarrow {PR}}\right)}\\&=\psi ((1-a-b)P+aQ+bR)\\&=(1-a-b)\psi (P)+a\psi (Q)+b\psi (R)\\&=\psi (P)+a{\overrightarrow {\psi (P)\psi (Q)}}+b{\overrightarrow {\psi (P)\psi (R)}}\\&=\psi (P)+a\varphi {\left({\overrightarrow {PQ}}\right)}+b\varphi {\left({\overrightarrow {PR}}\right)}\\&=\psi (P)+a\varphi (u)+b\varphi (v).\end{aligned}}$

Thus, we have

${}\varphi (au+bv)=a\varphi (u)+b\varphi (v)\,.$

To fact