Let $B$ be the algebraic group of upper triangular matrices. Then we know we can write
$$
B = MN,
$$
called the Levi decomposition, where $M$ is called a Levi subgroup such that $M^{\circ}$ (the maximal connected component of $M$ containing the identity) is reductive and $N$ is unipotent radical.
I was wondering what do $M$ and $N$ look like in this explicit case?
Thank you.
Best Answer
If you want to decompose it as a product of a nilpotent and a reductive group, it is $DN$ where $D$ is the diagonal invertible matrices and $N$ is the upper triangular matrices which values on the diagonal is $1$.