Diagonal Cartan subalgebra and roots of sl_n

By [L1], ${\mathfrak{s}\mathfrak{l}}_n(\mathbb{C})$ is semisimple.

$\mathfrak{h}$ is a Cartan subalgebra: it is abelian, hence nilpotent, and its normalizer is itself. Indeed if $X={\sum }_{ab}{x}_{ab}{E}_{ab}$ satisfies $[X,H]\in \mathfrak{h}$ for every diagonal traceless $H$, take $H=\mathrm{diag}(1,2,\dots ,n)-\frac{n+1}{2}I$, whose diagonal entries are pairwise distinct; then $[X,H]={\sum }_{ab}(h_b-h_a)x_{ab}{E}_{ab}$ has no off-diagonal component, so $(h_b-h_a)x_{ab}=0$ and hence $x_{ab}=0$ whenever $a\mathrm{\ne }b$. Thus $X$ is diagonal, and being in ${\mathfrak{s}\mathfrak{l}}_n$ it lies in $\mathfrak{h}$.

For $H=\mathrm{diag}(x_1,\dots ,x_n)\in \mathfrak{h}$ and a matrix unit $E_{ij}$ one computes $[H,E_{ij}]=(x_i-x_j)E_{ij}$; note $x_i-x_j$ depends only on $H$, so the functional ${\epsilon }_i-{\epsilon }_j$ on $\mathfrak{h}$ is well defined and $E_{ij}$ is a nonzero eigenvector for the eigenvalue $({\epsilon }_i-{\epsilon }_j)(H)$.

By steps 1.1 and 1.2, the root-space decomposition of Root-space decomposition applies. Step 1.3 exhibits, for every pair $i\mathrm{\ne }j$, the nonzero vector $E_{ij}\in {\mathfrak{g}}_{{\epsilon }_i-{\epsilon }_j}$; conversely every simultaneous $\mathfrak{h}$-eigenvector is a linear combination of those $E_{ij}$ whose indices give that functional, and the functionals ${\epsilon }_i-{\epsilon }_j$ for distinct ordered pairs are distinct while the diagonal matrices give the zero weight. Hence the roots are exactly the $n(n-1)$ functionals ${\epsilon }_i-{\epsilon }_j$, $i\mathrm{\ne }j$, with one-dimensional root spaces $\mathbb{C}{E}_{ij}$.