Every affine map is the sum of a linear transformation and a constant

Question

Definition: Let $\{p_0, p_1,...,p_m\}$ be an affine independent subset of $R^n$ and let $A$ denote the affine set it spans. An affine map $T: A \to R^k$ is a function satisfying $T(t_0p_0 +...+t_mp_m)= t_0T(p_0) + ...+ t_mT(p_m)$ whenever $t_0 + ...+ t_m = 1$.

If $T:R^n \to R^k$ is affine then there exists a linear transformation $\lambda: R^n \to R^k$ and a $y_0$ such that $T(x) = \lambda(x) + y_0$ for every $x \in R^n$.

Could someone provide a hint on how to reach the desired linear transformation? Thanks.

I have tried approaching it by assuming $\{ p_0, .... ,p_m\}$ is the affine independent set s.t $R^n$ is the affine set that it spans and then I considered the set $\{0, p_1 -p_0, ..., p_m - p_0\}$ clearly this set spans $R^n$ and is affine independent but We can’t use the property of the affine map here as this set differs from that of the original affine independent set and so we can’t bring out the $T(0)$ and set $y_0$ to be $T(0)$ by this approach.

Answer 1

$\{p_0, \ldots,p_m\}$ being affine independent means that the $m$ vectors $b_1 = p_1-p_0,\ldots, b_m = p_m-p_0$ are linearly independent. This implies $m \le n$. If $m = n$, these vectors form a basis of $\mathbb R^n$. If $m < n$, add vectors $b_{m+1},\ldots,b_n$ such that $b_1,\ldots,b_n$ becomes a basis of $\mathbb R^n$.

Define $\lambda(b_i) = 0$ for $m < i \le n$ and $\lambda(b_i) = T(p_i) - T(p_0) $ for $1 < i \le m$. This determines a unique linear map $\lambda : \mathbb R^n \to \mathbb R^k$. With $y_0 = T(p_0) - \lambda(p_0)$ we get for all $x = \sum_{i=0}^m t_ip_i \in A$

$$\lambda(x) = \lambda(\sum_{i=0}^m t_ip_i) = \lambda(\sum_{i=0}^m t_i(p_i-p_0) + (\sum_{i=0}^m t_i)p_0) = \lambda(\sum_{i=1}^m t_i(p_i-p_0) + p_0) \\ = \lambda(\sum_{i=1}^m t_i b_i + p_0) = \sum_{i=1}^m t_i \lambda(b_i) + \lambda(p_0) = \sum_{i=1}^m t_i (T(p_i) - T(p_0)) + \lambda(p_0) \\= \sum_{i=0}^m t_i (T(p_i) - T(p_0)) + \lambda(p_0) = \sum_{i=0}^m t_i T(p_i) - T(p_0) + \lambda(p_0) \\= T(\sum_{i=0}^m t_i p_i) - y_0 = T(x) - y_0 .$$

Answer 2

From your definition of an affine map it follows easyly that (*) $T(ux+vy)=u T(x)+v T(y)$ for all $x,y\in A$ and all $u,v\in\mathbb{R}$ such that $u+v=1$. Since $A$ is an affine subspace of $\mathbb{R}^n$ with $p_0\in A$ there is a vector subspace $V$ of $\mathbb{R}^n$ such that $A=p_0+V$. Let $\lambda\colon V\to \mathbb{R}^k$ be defined by $\lambda(x):=T(p_0+x)-T(p_0)$. Then $\lambda(0)=0$ and $$\lambda(\frac12 x+\frac12y)=T(p_0+\frac12 x+\frac12 y)-T(p_0)=T(\frac12(p_0+x)+\frac12(p_0+y))-T(p_0)=\frac12 T(p_0+x)+\frac12 T(p_0+y)-T(p_0)=\frac12(\lambda(x)+T(p_0))+\frac12(\lambda(y)+T(p_0))-T(p_0)=\frac12\lambda(x)+\frac12\lambda(y)$$ This for $y=0$ results in $\lambda(\frac12 x)=\frac12\lambda(x)$. Together with the result above we get $\frac12\lambda(x+y)=\lambda(\frac12(x+y))=\lambda(\frac12x+\frac12 y)=\frac12\lambda(x)+\frac12\lambda(y)$, i.e., $\lambda(x+y)=\lambda(x)+\lambda(y)$ for all $x,y\in V$. So $\lambda$ satisfies one of the properties of linearity. Let $u$ be real and $x\in V$. Then $\lambda(ux)+T(p_0)=T(p_0+u x)=T((1-u) p_0+u (x+p_0))=(1-u)T(p_0)+uT(x+p_0)=T(p_0)+u\lambda(x)$, i.e., $ \lambda(ux)=u\lambda(x)$. So there is a linear map $\lambda\colon V\to\mathbb{R}^k$ such that $T(y)=\lambda(y-p_0)+T(p_0)$.

Note that there is a linear extension $\Lambda\colon\mathbb{R}^n\to\mathbb{R}^k$ of $\lambda$. Using this we get that $T(y)=\Lambda(y)+y_0$ for all $y\in A$ where $y_0=T(p_0)-\Lambda(p_0$.

This characterization of affine maps also works for all affine subspace of arbitrary real vector spaces with arbitrary real vector spaces as codomain, if affinity of a map is defined by (*).