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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?

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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.)

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