The composition of two homotheties is either a homothety or translation

Question

Let $f$ and $g$ be two homotheties that don't have the same center.

What kind of affine transformation is their transformation?

I know it can be either a homothety or translation, but I don't know how to prove one of these alternatives.

Answer 1

@Tsemo Aristide I am sorry but I am uncomfortable with your presentation, because I don't see in which set all this happen. Are $a$ and $b$, real numbers, complex numbers, affine tranforms?

For me, there are two choices:

• assimilate $\mathbb{R^2}$ with $\mathbb{C}$ and compose transformations of the form $z'=az+b$ as you have done.

• take the general case, valid for any $\mathbb{R^n}$, i.e. consider affine transforms of the form (for the homothety with center $X_0 (x_0,y_0)$ and ratio $a$):

$$\begin{bmatrix}x'\\y'\end{bmatrix}-\begin{bmatrix}x_0\\y_0\end{bmatrix}=\begin{bmatrix}a&0\\0&a\end{bmatrix}\left(\begin{bmatrix}x\\y\end{bmatrix}-\begin{bmatrix}x_0\\y_0\end{bmatrix}\right)$$

or, synthetically:

$$X'-X_0=aI(X-X_0) \ \ with \ \ a \neq 1 \ \ (form 1)$$

or

$$h(X)=X'=aX+B \ \ with \ \ a \neq 1 \ \ (form 2) $$

for a certain $B$. The expression of $B$ is in fact unimportant, because the homothety center can always be retrieved as the fixed point of $h$ (the point $X_0$ such that $X_0=aX_0+B$, i.e., $X_0=\frac{1}{1-a}B$. As we can write (form 2) in this way:

$$(X'-\frac{1}{1-a}B)=a(X-\frac{1}{1-a}B)$$

(form (1)) we can say that the set of homotheties is exactly the set of transformations:

$$h(X)=X'=aX+B \ \ \text{for any} \ \ a \neq 1 \ \ and \ \ B\in\mathbb{R^n} $$

Then, composing $h_1$ and $h_2$, given by $h_1(X)=a_1X+B_1$ and $h_2(X)=a_2X+B_2$ is

$$h_2\circ h_1(X)=a_2(a_1X+B_1)+B_2=(a_2a_1)X+(a_2B_1+B_2)$$

Thus,

• either $a_2a_1 = 1$ and we obtain a translation $X'=X+B_3$.

• or $a_2a_1\neq 1$ and we have an homothety with ratio $a_2a_1$.

Answer 2

It depends. Write $f(x)=ax, g(x)=bx+y, fg(x)=abx+ay$. If $ab=1$, $f\circ g$ is a translation of direction $ay$, otherwise it is an homothetic map.