Examples of root, parabolic, and borel subgroups corresponding to roots I'm interested in seeing a few examples of root, parabolic, and Borel subgroups given a specific reductive group $G$.  Here is what I know.
Let $G$ be a reductive algebraic group over an algebraically closed field $k$ of characteristic $p$.  Fix a maximal torus $T\subset G$, and let $T$ act on Lie$(G)=\mathfrak{g}$ via the adjoint action.  Then $\mathfrak{g}$ decomposes as a direct sum of eigenspaces corresponding to this action.  The eigenvalues are the roots in the group of characters of the torus $T$.  Let the roots of $G$ with respect to $T$ be denoted by $\Phi$.  Choose a set of simple roots $\Pi\subset\Phi$.
For each root $\alpha$, we can uniquely define a root subgroup $U_{\alpha}$ of $G$ as the image of a certain one-parameter subgroup $\mathbb{G}_a\to G$.  Furthermore, given a subset $J\subset\Pi$ of simple roots, we can define $P_J=\langle T,U_{\alpha}|\alpha\in J\rangle$.  Finally, there is exactly one Borel subgroup which contains the root subgroups corresponding to all positive roots (which depend on the choice of $\Pi$).
Is all of this correct?  If so, it doesn't mean much to me without seeing an example or two worked through.  What are examples of these subgroups in the case that $G=GL_n$ and $T$ is the diagonal subgroup?  If this is too much to type out here, I would appreciate a link to any online notes that work through this example thoroughly.  It would be nice to see how $T$ determines the set of roots $\Phi$, what the root subgroups are for an arbitrary root, what the parabolic subgroups are for subsets of a choice of simple roots, and what the corresponding Borel subgroup is.
 A: I am also a beginner in this field. The following is an example.
For $G=GL_n$, $T=\{\mathrm{diag}(c_1,\cdots c_n),c_i\in G_m\}$ the set of diagonal matrices in $G$. $T\simeq G_m^n$ is a torus of $G$.


*

*Consider the character group $X^*(T)$. It consists a natural basis
$$
\epsilon_j:T\rightarrow G_m,\quad \epsilon_j(\mathrm{diag}(c_1,\cdots c_n))=c_j
$$
Every characters of $T$ is a $\mathbb{Z}$-linear combinatinon of $\epsilon_j$.

*$Lie(G)=\mathfrak{gl}_n$, the set of $n\times n$ matices. It has a natrual basis $E_{ij}$, the matrix with $1$ at $(i,j)$ and $0$ otherwise.

*For $t=\mathrm{diag}(c_1,\cdots c_n)\in T$, consider the adjoint action on $E_{ij}$.
\begin{eqnarray*}
\mathrm{Ad}t.E_{ij}=\begin{cases}E_{ij},&\quad i=j\\ (c_ic_j^{-1})E_{ij}=(\epsilon_i-\epsilon_j)(t)E_{ij}&\quad i\neq j\end{cases}
\end{eqnarray*}
It shows that $E_{i,i}$ are in weiht zero space, and $E_{ij},i\neq j$ are non-zero weight space of weight $\alpha_{ij}=\epsilon_i-\epsilon_j$.

*Thus the root system of $G=GL_n$ is
$$\Phi=\{\pm\alpha_{ij}=\pm(\epsilon_i-\epsilon_j),1\leq i< j\leq n\}.$$
One can choose a set of positive roots 
$$\Phi^+=\{\alpha_{ij}=(\epsilon_i-\epsilon_j),1\leq i< j\leq n\}.$$
and the set of simple roots
$$\Delta=\{\alpha_{i,i+1}=(\epsilon_i-\epsilon_{i+1}),1\leq i< j\leq n\}.$$
It is easy to see that elements in $\Phi$ are $\mathrm{Z}$-linear combinition of $\Delta$, and elements in $\Phi^+$ are with non-negative coefficients.
A: We cosider $X_*(T)=\{\lambda^\vee:G_m\rightarrow T\}$, the set of $1$-parameters subgroups.


*

*It is easy to see that $X_*(T)$ is a free $\mathbb{Z}$-module with generators
\begin{eqnarray}
\epsilon_j^\vee:G_m\rightarrow T,\quad c\mapsto \mathrm{diag}(1,\cdots c,\cdots 1)
\end{eqnarray}

*Consider the composition:
\begin{eqnarray}
\epsilon_j\circ\epsilon_i^\vee:G_m\rightarrow T\rightarrow G_m
\end{eqnarray}
It is easy to see that $\epsilon_j\circ\epsilon_i^\vee=0$ if $i=j$, and $\epsilon_j\circ\epsilon^\vee_j=1$ is the identity map on $G_m$.
Thus we have a $\mathbb{Z}$-binlinear map
\begin{eqnarray}
X^*(T)\times X_*(T)\rightarrow \mathbb{Z},\quad (\lambda,\mu^\vee)\mapsto\lambda\circ\mu^\vee
\end{eqnarray}
and $\{\epsilon_i\}$ and $\{\epsilon_i^\vee\}$ are the dual $\mathrm{Z}$-basis.

*Given $\alpha_{ij}=\epsilon_i-\epsilon_j$. Consider $\alpha_{ij}^\vee=\epsilon_i-\epsilon_j$. It is an element in $X_*(T)$ and one has
$$\alpha_{ij}\circ\alpha_{ij}^\vee= \epsilon_i\circ\epsilon_i^\vee+(-\epsilon_j)\circ(-\epsilon_j^\vee)=2.$$
Thus $\alpha_{ij}^\vee$ are the coroots of $\alpha_{ij}$.
The set of corroots are
$$
\Phi^\vee=\{\pm\alpha_{ij}^\vee=\pm(\epsilon_i-\epsilon_j),1\leq i<j\leq n\}
$$
A: Now we have the root datum of $G=GL_n$.
We also have chosen a set of positive roots $\Phi^+$ as above.
For each $\alpha_{ij}\in\Phi^+$, as the above argument, the corresponding root space are 
$$\mathfrak{gl_n}_{\alpha_{ij}}=\mathbb{C}E_{ij}.$$
Roughly speaking,
the torus $T$ and all positive root spaces corresponds to upper triangular matrix, which is just the Borel subgroup.
