Show that $g^{-1} \circ f \circ g$ is a translation, if $g$ is a rigid motion and $f$ is a translation

Question

Question:

Let $g: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a rigid motion, and let $f: \mathbb{R}^n \rightarrow \mathbb{R}^n$ be a translation.

Show that $g^{-1} \circ f \circ g$ is also a translation.

My attempt:

I know that for any finite dimensional vector space over $\mathbb{R}$, any rigid motion $g(x)$ can be written as: $g(x) = f_{g(0)} \circ T(x)$, where:

• $f_{g(0)}$ is a translation given by $f_{g(0)}(w) = w + g(0)$
• $T$ is an orthogonal and linear operator given by $T(x) = f_{-g(0)} \circ g(x)$

We note that both $f_{g(0)}$ and $T$ are invertible, and the inverse of $f_{g(0)}$ is $f_{-g(0)}

So then $g^{-1} \circ f \circ g = T^{-1} \circ f_{-g(0)} \circ f \circ f_{g(0)} \circ T$

I assume it's safe to say here that translations commute, so the above simplifies to:

$T^{-1} \circ f \circ T$

I'm not sure if my working up to this point has been correct, and I'm also not sure how to proceed from here. Any suggestions?

Thank you!

Answer 1

Let $h = T^{-1}\circ f\circ T$. Observe that $f(v) = v + f(0)$ and that $h(0) = T^{-1}(f(0))$. We have

$$\begin{align} h(v) - h(0) &= T^{-1}\big(\,f(T(v))\,\big) - T^{-1}(f(0)) \\&= T^{-1}\big(f(T(v)) - f(0)\big) \tag{$T^{-1}$ is linear} \\&= T^{-1}\big(T(v) + f(0) - f(0)\big) \tag{$f$ is a translation} \\&= T^{-1}\big(T(v) \big) = v, \end{align}$$

that is, $h(v) = v + h(0)$. In other words, $h$ is a translation.

Answer 2

Suppose $f(x) = x + v$ is translation by an arbitrary vector $v$. Explicitly calculate $(g^{-1} \circ f \circ g)(x)$ using the formula for $f$ and the fact that $g$ and $g^{-1}$ are linear.