Help with a proof in Humphreys

Question

I was reading a proof but failed to see how the underlined step goes.
How come ad $y = r($ad $s)$? That means $$r(a_i-a_j)e_{ij} = f(a_i-a_j)e_{ij} =ad\;y\;(e_{ij}) = r(ad\; s\;(e_{ij})) =r ((a_i-a_j)e_{ij})$$ but how can this be true?

Answer 1

The basis $(e_{ij})$ diagonalises both $\mathrm{ad}(y)$ and $\mathrm{ad}(s)$, and, by construction, the polynomial $r$ maps the diagonal values of $\mathrm{ad}(s)$ (relative to said basis) to the diagonal values of $\mathrm{ad}(y)$ (in the same basis.)