What is the most common/convenient set of infinitesimal generators of $GL^{+}(n,R)$ as a Lie group?
We can write any matrix element of $GL^{+}(n,R)$ in the form $\exp(t A_{n\times n})$ where the basis for $A_{n\times n}$ are identified as the infinitesimal generators.
Naively, I think that one possible set could be those with only one entry being $1$ and all other entries are $0$. For example, for $GL^{+}(n,R)$ there are $9$ generators and some of them are
$\begin{pmatrix}1 & 0 &0 \\0 & 0 &0\\0 & 0 &0\end{pmatrix}, \; \begin{pmatrix}0 & 1 &0 \\0 & 0 &0\\0 & 0 &0\end{pmatrix}, \; \begin{pmatrix}0 & 0 &1 \\0 & 0 &0\\0 & 0 &0\end{pmatrix},\; \begin{pmatrix}0 & 0 &0 \\1 & 0 &0\\0 & 0 &0\end{pmatrix}, \cdots$
For the Lie algebras we have $$ {\rm Lie}(GL_n^+(\mathbb{R}))\cong {\rm Lie}(GL_n(\mathbb{R}))\cong \mathfrak{gl}_n(\mathbb{R}). $$ This is just the vector space of $n\times n$-matrices with Lie bracket $[A,B]=AB-BA$. The matrices $E_{ij}$ with zero entries except for entry $1$ at position $(i,j)$ form a Lie algebra basis, the "infinitesimal generators".