How to find root subgroups $\newcommand{\GL}{\text{GL}}\newcommand{\diag}{\text{diag}}$For $G = \GL_n(k)$ let $B$ be the upper triangular matrices and $T$ be the diagonal matrices in $G$. In this case I understand that the positive roots $\Phi^+ = \{e_{ij}\}_{i< j}$ are given by
$$
e_{ij}(\diag(\lambda_1\dots \lambda_n)) = \lambda_i\lambda_j^{-1}.
$$
for $i<j$.
For each root $e_{ij}\in \Phi^+$ we get the root subgroup which I have been told is
$$
U_{e_{ij}} = \{I + \lambda E_{ij}\mid \lambda \in k\}.
$$
($E_{ij}$ is the matrix with zeros all over except a $1$ in the $ij$th entry.)
I would like to see how one actually computes these root subgroups. How was it determined that $U_{e_{ij}}$ looks like above? What is the definition of a root subgroup? How, using the definition, can one actually compute these root subgroups?
 A: Depending on your definitions, the "computation" of the root subgroups for $GL_n$ is mostly an unpacking of those definitions once the characters and positive roots have been determined.  Disclaimer: I will be using the definitions and conventions set out in Malle and Testerman's Linear Algebraic Groups and Finite Groups of Lie Type.  
First, let me fix some notation (in addition to yours).  If $\alpha$ is a root of $G$, let $T_\alpha = (\ker \alpha)^\circ \leq T$ be the connected component of $\ker (\alpha: T \to k^\times)$, $C_\alpha = C_G(T_\alpha)$ be the centralizer of $T_\alpha$ in $G$, and $G_\alpha = [C_\alpha, C_\alpha]$ be the commutator subgroup of $C_\alpha$.  Let $B'$ be a Borel subgroup in $G_\alpha$.  (There is some ambiguity here, which is resolved in theory by considering the Lie algebra $\mathfrak{g}_\alpha$ or $\mathfrak{g}_{-\alpha}$ obtained -- see below.  But if we are dealing with the positive roots, then the correct choice of Borel subgroup here is $B' = B \cap G_\alpha$)  Then the root subgroup you are looking for is defined to be $U_\alpha = B'_u$, the subgroup of unipotent elements in $B'$.  
In your specific case $G = GL_n(k)$, let $\alpha = e_{12}$.  (The computation for the other roots in similar, but more difficult to TeX.)  Also, assume $k \neq \mathbb{F}_2$.  We have
\begin{align*}
T_\alpha &= \{\operatorname{diag}(\lambda_1, \ldots, \lambda_n): \lambda_1 = \lambda_2\} \\
C_\alpha &= \left\{\begin{pmatrix} * & * & 0 & \cdots & 0 \\ * & * & 0 & \cdots & 0 \\
0 & 0 & * & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & * \end{pmatrix} \in GL_n(k) \right\} \\
G_\alpha &= \left\{\begin{pmatrix} * & * & 0 & \cdots & 0 \\ * & * & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} \in SL_n(k) \right\}
\end{align*}
Notice that at this point we have basically reduced the problem to $SL_2(k)$. (This is common in other computations too.)  Continuing on with the computation,
\begin{align*}
B' &= \left\{\begin{pmatrix} * & * & 0 & \cdots & 0 \\ 0 & * & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix} \in SL_n(k) \right\} \\
U_\alpha &= \left\{\begin{pmatrix} 1 & * & 0 & \cdots & 0 \\ 0 & 1 & 0 & \cdots & 0 \\
0 & 0 & 1 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \end{pmatrix}: * \in k \right\} 
\end{align*}
This is exactly the group $\{I + \lambda E_{12}\}$ you were looking for.  
Some other remarks: the root subgroup can be defined in slightly different ways.  For example, $U_\alpha$ is the unique one-dimensional connected unipotent subgroup of $G$ normalized by $T$ with Lie algebra $\operatorname{Lie}(U_\alpha) = \mathfrak{g}_\alpha$, where $\mathfrak{g}_\alpha$ is the vector subspace corresponding to the root $\alpha$ in the root space decomposition of the Lie algebra $\mathfrak{g}$.  From this perspective, $U_\alpha$ is also the unique one-parameter group in $G$ given by a morphism $u_\alpha: k \to G$ satisfying $tu_\alpha(c)t^{-1} = u_\alpha(\alpha(t)c)$ for all $t \in T$ and $c \in k$.  If you are willing to accept these facts, you can also just check that the root subgroups you are given have the required properties.  
