Composition of homotheties does not look homothetic

Question

Given an affine space $X$, let us define homotheties $h_1,h_2$ as

$$ h_1(x) = c_1 + \lambda_1\overset\longrightarrow{c_1x} $$

$$ h_2(x) = c_2 + \lambda_2\overset\longrightarrow{c_2x} $$

for some $\lambda_1, \lambda_2 \in \mathbb R$, $c_1, c_2 \in X$.

The composition $h_2 \circ h_1$ can thus be written as $$ h_2 \circ h_1(x) = c_2 + \lambda_2\overset\longrightarrow{c_2x'} $$ where $x'=h_1(x)$.

Now, if we consider $\overset\longrightarrow{c_2x'}=\overset\longrightarrow{c_2c_1}+\lambda_1\overset\longrightarrow{c_1x}$, then $$ h_2 \circ h_1(x) = c_2 + \lambda_2(\overset\longrightarrow{c_2c_1}+\lambda_1\overset\longrightarrow{c_1x})=c_2 + \lambda_2\overset\longrightarrow{c_2c_1}+\lambda_2\lambda_1\overset\longrightarrow{c_1x} $$ This looks like it is only a homothety when $c_2 + \lambda_2\overset\longrightarrow{c_2c_1}=c_1$, and not a translation since the vector depends on $x$. However, the composition of two homotheties should be either homothetic or a translation.

Where is the mistake?

Answer 1

You can check that: $$ c_2 + \lambda_2\overset\longrightarrow{c_2c_1}+\lambda_2\lambda_1\overset\longrightarrow{c_1x}= c_3 +\lambda_2\lambda_1\overset\longrightarrow{c_3x}, \quad\hbox{where}\quad c_3=c_2-{\lambda_2\over\lambda_2\lambda_1-1}\overset\longrightarrow{c_1c_2}. $$ I found the center $c_3$ of the resulting homothety as the solution of the equation $h_2 \circ h_1(x)=x$.

Of course this doesn't work if $\lambda_2\lambda_1=1$. In that case the result is just a translation: $$ h_2 \circ h_1(x)=c_2 + \lambda_2\overset\longrightarrow{c_2c_1}+\overset\longrightarrow{c_1x}= x+(\lambda_2-1)\overset\longrightarrow{c_2c_1}. $$