Computing $Z(G)$ Let $G=\{A \in GL_n( \mathbb{C}) : a_{ii}=1$ and $a_{ij}=0$ when $i<j\}$. I see that $G$ is a group, a subgroup of $GL_n(\mathbb{C})$. My question is, what is $Z(G)$ and how do we compute it?
When $n=2$, I see that $G$ will be abelian so of course we just have $Z(G)=G$.
When $n=3$, however, it is not abelian. We have $\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & e & f \\ 0 & 1 & g \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & e & f \\ 0 & 1 & g \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix}$ if and only if $ag=ce$. 
So, $\begin{bmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1 \end{bmatrix} \in Z(G)$ iff the above is true for any choice of $e,g$ (i.e. iff $a=c=0$). 
Thus $Z(G)$ is merely the set $\{\begin{bmatrix} 1 & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} : b \in \mathbb{C} \}$.
But what do we do for $n=4,5,etc.$?
Bonus question: Will our result for $Z(G)$ depend on the fact that we are using the field $F= \mathbb{C}$? My guess is that it will work for any field of $char(F)=0$, but for general fields I have some doubts it will be the same.
 A: For the $n \times n$ case, every element of your group can be written in the form
$$
M = I + \sum_{i < j} a_{ij} E_{ij}
$$
where $E_{ij} = e_i e_j^T$ is the matrix with a $1$ in the $(i,j)$ entry.  $e_i$ denotes the $i$th standard basis vector.
First, note that $M$ will be in the center of $G$ if and only if $ME_{ij} = E_{ij}M$ for all $i<j$ (which is true if and only if $M(I + E_{ij}) = (I + E_{ij})M$ for all $i<j$).  We note also that for any $i,j,p,q,$ we have
$$
E_{ij}E_{pq} = e_ie_j^Te_pe_q^T = (e_j^Te_p)e_ie_q^T = 
\begin{cases}
E_{iq} & j=p\\
0 & j\neq p
\end{cases}.
$$
Now, for a fixed $1 \leq p<q\leq n$ and upper-triangular $M$, we compute
$$
E_{pq}M = E_{pq} + \sum_{i < j} m_{ij} E_{pq}E_{ij}
= E_{pq}+ \sum_{q<j}m_{qj} E_{pq}E_{qj}\\
= E_{pq}+ \sum_{j>q}m_{qj} E_{pj}
= e_p(e_q + m_{q,q+1}e_{q+1} + \cdots + m_{q,n}e_n)^T
$$
And similarly, 
$$
ME_{pq} = (m_{1,p}e_1 + \cdots + m_{p-1,p}e_{p-1} + e_p)e_q^T.
$$
Now, what conditions on the entries $m_{ij}$ ensure that $E_{qp}M = ME_{pq}$? Note that these equations place no constraints on the entry $m_{1n}$, and verify that all other entries $m_{ij}$ (for $1\leq i < j \leq n$) appear in the equations above for some choice of $p$ and $q$.  
In other words, we should find that $M$ lies in the center of $G$ if and only if all entries above the diagonal are zero, except possibly for $m_{1n}$.  The characteristic of the underlying field is irrelevant since none of these calculations required division.
