Exponentiation of the Euclidean algebra $\text{e}(n)$

Question

We know that the Euclidean group $E(n)$ is isomorphic to the set of real matrices of the form $$ E=\begin{bmatrix} A &a\\ 0 &1 \end{bmatrix} $$ where $A\in O(n)$ and $a\in \mathbb{R}^n$. It's also clear that it's necessary for matrices in the Lie algebra $\text{e}(n)$ to be of the form $$ e=\begin{bmatrix} \omega &v\\ 0 &0 \end{bmatrix} $$ where $\omega \in \text{o}(n)$ and $v\in \mathbb{R}^n$. Then what is the exponentiation of $e$, i.e., what is $\exp(et)$ for $t\in \mathbb{R}$?

Answer 1

Since $$ e=\begin{bmatrix} \omega &v\\ 0 &0 \end{bmatrix} ~, $$ prove $$ t^n e^n=\begin{bmatrix} t^n \omega^n &t^n \omega^{n-1}v\\ 0 &0 \end{bmatrix} $$ and the formal expansion of the exponential sums to $$ E=\exp (te) = \begin{bmatrix} \exp (t\omega) & \frac{1}{\omega} ~(\exp (t\omega) - 1\!\! 1 )~ v \\ 0 &1 \end{bmatrix} ~. $$ Of course, no actual inverse of ω is required to exist, since the formal expansion of the numerator of the matrix acting on v is linear in ω.