Let $A:\mathbb{R}^n\to \mathbb{R}^m $ is affine function iff $ A(tx+(1-t)y)=tA(x)+(1-t)A(y)$ $\forall t\in \mathbb{R},x,y\in \mathbb{R}^n$.
I can prove that
$A$ is affine then it satisfies $A(tx+(1-t)y)=tA(x)+(1-t)A(y) \ \forall t\in \mathbb{R},x,y\in \mathbb{R}^n$.
But I could not prove converse ?
A mapping $A:\mathbb{R}^n\to \mathbb{R}^m$ is said to be affine if there exist linear map $B:R^n\to R^m$ and element $b\in R^m$ such that $A(x)=B(x)+b$
Any Hint will be appreciated .
Thanks a lot
Define $B(x):=A(x)-A(0)$. Then show that $B$ is linear: $$ B(t x) = A(tx)-A(0) = A(tx+(1-t)0)-A(0) = tA(x)+(1-t)A(0)-A(0) = t(A(x)-A(0)) = tB(x) $$ and $$ B(x+y) = 2B(\frac x2+\frac y2)= 2(\frac 12B(x)+\frac 12B(y))=B(x)+B(y). $$ Then set $b:=A(0)$.