Here the "the conditions that define" the Lie group are $a_{ij}=0, i<j$ differentiate that you obtain again $a_{ij}=0$ with no other constraints.
Generally, you can't have an invertibility constraint in a Lie algebra since it is a vector space.
The exponential map $\exp\colon \mathfrak{n}^+\rightarrow N$ from the nilpotent Lie algebra of all stricly upper-triangular matrices to the nilpotent group of all upper uni-triangular matrices is given by polynomial maps with inverse given by matrix logarithm. Hence $\exp$ is bijective in this case. This is no longer true for upper-triangular matrices. The question when $\exp$ is injective is easier than the question when $\exp$ is surjective. For injectivity we have very nice criteria, given for example at this MSE-question:
If $G$ is a real Lie group with Lie algebra $\mathfrak{g}$, then the following are equivalent:
- $\exp$ is injective
- $\exp$ is bijective
- $\exp$ is a real analytic diffeomorphism
- $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ as subalgebra of a quotient.
- $G$ is solvable, simply connected, and $\mathfrak{g}$ does not admit $\mathfrak{e}$ or $\tilde{\mathfrak{e}}$ as subalgebra
Here $\mathfrak{e}$ is the 3-dimensional Lie algebra with basis $(H,X,Y)$ and bracket $[H,X]=Y$, $[H,Y]=-X$, $[X,Y]=0$. It is isomorphic to the Lie algebra of the group of isometries of the plane. Its central extension $\tilde{\mathfrak{e}}$ is defined as the 4-dimensional Lie algebra defined by adding a central generator $Z$ and the additional nonzero bracket $[X,Y]=Z$.
Concerning surjectivity the situation is more complicated, see
On surjectivity of exponential map for Lie groups, or Exponential map is surjective for compact connected Lie group.
So, for 1. Yes, $\exp$ is surjective, because every upper-triangular real matrix with positive diagonal entries is invertible and is the square of an invertible matrix, see here.
For 2.) Yes.
Best Answer
The Lie algebra $\mathfrak{t}_n(K)$ of upper-triangular matrices cannot be nilpotent, because it has a non-nilpotent solvable Lie subalgebra of dimension $2$, e.g., generated by $E_{12}$ and $E_{11}-E_{22}$. Hence it cannot be nilpotent.