Let $UT_1(n, \mathbb R)$ denote the set of all upper triangular real matrices with diagonal equal to $1$. This is a Lie group. I am trying to determine its Lie algebra.
Please can you tell me if my thoughts are correct?
If $A=A(t)$ is a differentiable curve $A: (-1,1)\to UT_1(n, \mathbb R)$ with $A(0) = I$ then its diagonal is constant and since we take the derivative termwise, $A'$ has a zero diagonal.
The remaining upper triangle consists of any elements. I think this because the Lie algebra of the general linear group is all matrices. Now we can take the subgroup of triangular matrices (that's the set of invertible triangular matrices) and by the same reasoning its Lie algebra should be all triangular matrices.
Hence,
$$ \mathfrak{ut}_1 = \{ A \in M_n(\mathbb R) \mid A \text{ is upper triangular with zero diagonal }\}$$