Type $A_n$ - $\mathfrak{sl}_{n+1}(\mathbb{C})$: Traceless complex $(n+1) \times (n+1)$ matrices with bracket $[X,Y] = XY - YX$.

Cartan subalgebra: diagonal matrices $\text{diag}(h_1, \ldots, h_{n+1})$ with $\sum h_i = 0$

Root vectors: $E_{ij}$ (matrix with 1 in position $(i,j)$, zero elsewhere) for $i \neq j$

Simple roots: $\alpha_i = e_i - e_{i+1}$ for $i = 1, \ldots, n$

Type $B_n$ - $\mathfrak{so}_{2n+1}(\mathbb{C})$: Skew-symmetric $(2n+1) \times (2n+1)$ matrices preserving the standard bilinear form.

Can be realized as block matrices: $\begin{pmatrix} A & u & v^T \\ -v & 0 & u^T \\ w^T & -u & -A^T \end{pmatrix}$ where $A$ is $n \times n$, $u, v, w$ are $n \times 1$.

Root system has short roots $\{\pm e_i\}$ and long roots $\{\pm e_i \pm e_j\}$.

Type $C_n$ - $\mathfrak{sp}_{2n}(\mathbb{C})$: Matrices preserving the symplectic form $J = \begin{pmatrix} 0 & I_n \\ -I_n & 0 \end{pmatrix}$: $\mathfrak{sp}_{2n} = \{X : X^T J + JX = 0\}$

Block form: $\begin{pmatrix} A & B \\ C & -A^T \end{pmatrix}$ where $B, C$ are symmetric $n \times n$ matrices.

Root system has long roots $\{\pm 2e_i\}$ and short roots $\{\pm e_i \pm e_j\}$.

Type $D_n$ - $\mathfrak{so}_{2n}(\mathbb{C})$: Skew-symmetric $2n \times 2n$ matrices. Root system $\{\pm e_i \pm e_j : i \neq j\}$ has all roots the same length.

The Dynkin diagram fork reflects the even-dimensional spinor representations' special properties (in $D_4$, this becomes triality).

Type $G_2$ - Octonion derivations: The automorphism algebra of the octonions $\mathbb{O}$ gives $G_2$. Alternatively, construct from $\mathfrak{sl}_3$ by adding derivations of a cubic form, yielding 14-dimensional algebra.

Root system fits in a plane with 6 short and 6 long roots (length ratio $\sqrt{3}$).

Types $E_6, E_7, E_8$ - Chevalley construction: Start with the Cartan matrix, define generators $\{e_i, f_i, h_i : i = 1, \ldots, n\}$ satisfying Serre relations: $[h_i, e_j] = a_{ij} e_j, \quad [h_i, f_j] = -a_{ij} f_j$ $[e_i, f_j] = \delta_{ij} h_i$ $(\text{ad}_{e_i})^{1-a_{ij}}(e_j) = 0, \quad (\text{ad}_{f_i})^{1-a_{ij}}(f_j) = 0$

This presentation constructs the algebra purely from Dynkin diagram data.