Upper Triangular Matrices is a lie group

Question

I'm trying to prove that this is a lie group and compute its tangent space at the identity. I would try to show its a closed subgroup of $GL(n)$ by showing its the preimage of a closed set under a continuous map on $GL(n)$, but doing it this way doesn't help compute the tangent space at the identity (I'm also struggling to actually find such a map). Any suggestions?

Answer 1

I believe you talking about the invertible upper triangular matrices $IUT(n)$. To show that its is a Lie group, you can to show it is closed in $Gl(n)$, it is defined by $a_{ij}=0, i>j$. So it is closed.

Let $UT(n)$ be the space of upper triangular matrices, it is a vector spaces and $IUT(n)$ is an open subspace of $UT(n)$, this vector space induces on $IUT(n)$ a structure of a manifold such that the multiplication, the inverse map are continuous for the topology, so it is a structure of Lie group and the tangent space at any point is just $UT(n)$.