The a2 serre relations

Set $z=[e_1,e_2]$. The two positive relations say $[e_1,z]=[e_2,z]=0$. Hence the span of $e_1,e_2,z$ is a Lie subalgebra containing the positive generators and equals the positive half. The same argument gives at most three dimensions for the negative half. Since $\det A=3$, the Cartan has dimension two, so F1 gives $\dim \mathfrak{g}\le 8$.

Take $e_1=E_{12}$, $e_2=E_{23}$, $f_1=E_{21}$, $f_2=E_{32}$, $h_1=E_{11}-E_{22}$ and $h_2=E_{22}-E_{33}$. The identity $[E_{ab},E_{cd}]={\delta }_{bc}{E}_{ad}-{\delta }_{da}{E}_{cb}$ gives $[e_1,e_2]=E_{13}$, $[f_2,f_1]=E_{31}$, $[e_i,f_j]={\delta }_{ij}{h}_i$, and zero second brackets $[E_{12},E_{13}]=[E_{23},E_{13}]=0$, with the analogous negative zeros. For a diagonal $h=\mathrm{diag}(t_1,t_2,t_3)$, $[h,E_{ab}]=(t_a-t_b)E_{ab}$. Thus ${\alpha }_1=t_1-t_2$, ${\alpha }_2=t_2-t_3$ give $({\alpha }_j(h_i){)}_{ij}=A$. All presentation relations hold, so F1 gives a homomorphism.

The six off-diagonal matrix units and the two displayed diagonal matrices are independent and span all traceless matrices. Step 1.2 therefore gives a surjection onto an eight-dimensional algebra; step 1.1 makes it an isomorphism. The diagonal commutator formula assigns the six weights stated in the example to these six units; the diagonal subspace has weight zero. Thus the root list is exhaustive, with each root multiplicity one.