I am having difficulties understanding why the roots of $\mathfrak{su}(N)$ algebra are $\alpha_{ij}=e_i-e_j$. This is (roughly) what my professor said:
We define the ladder operators in $\mathfrak{su}(N)$ as the matrices $E_{ij}$ such that they have a $1$ in the $(i,j)$-position and $i\neq j$. Given $H=\operatorname{diag}(\lambda_1,\dots,\lambda_N)$ we have $[H,E_{ij}]=(\lambda_i-\lambda_j)E_{ij}$. Then, if we take the elements of Cartan subalgebra to be $E_{ii}-\frac1N I_N$ (where $I_N$ is the identity matrix) we get that the roots are $$\alpha_{ij}=e_i-e_j$$ where $i\neq j$ so there are $N(N-1)$ of them.
I understand $[H,E_{ij}]=(\lambda_i-\lambda_j)E_{ij}$ (example on WA) but I don't get how $$[E_{kk}-\frac1N I_N,E_{ij}]$$ would give those roots.
P. S. I know that in general $[H_i,E_\alpha]=\alpha_i\,E_\alpha$.
I tried to understand it this way: ($I_N$ commutes with $E_{ij}$) $$[E_{kk},E_{ij}]=([E_{kk}]_{ii}-[E_{kk}]_{jj})E_{ij}=(\delta_{ki}-\delta_{kj})E_{ij}$$ since $E_{kk}$ has a $1$ in $kk$ entry, we have $[E_{kk}]_{ii}=\delta_{ki}$.
Then, in analogy with $[H_i,E_\alpha]=\alpha_i\,E_\alpha$ we found that $\delta_{ki}-\delta_{kj}$ is the $k$-th entry of the root $\alpha_{ij}$ associated with $E_{ij}$. Therefore
$$[\alpha_{ij}]_k=\begin{cases}0&k\neq i,j\\1&k=i\\-1&k=j\end{cases}$$ so $\alpha_{ij}=(0,\dots,0,1,\dots,-1,0,\dots,0)$ where the $1$ and $-1$ are in $i$-th entry and $j$-th entry respectively. This is equivalent to say $\alpha_{ij}=e_i-e_j$