In the book Introduction to Lie Algebras by Erdmann and Wildon, one finds the following result: given $V$ a complex vector space and $L$ a Lie subalgebra of $\mathfrak{gl}(V)$, if $tr(xy) = 0\,\, \forall x\in L', y \in L, $ then $L$ is solvable. At the end of the proof one reads the following:
$$p(ad(x)) = \overline {ad(d)} = ad(\bar d) \quad (1)$$
I have problems in understanding the both equalities. With $p$ is meant a polynomial in $C[X]$ with $p(x) =\overline d$, $ad$ is the adjoint homomorphism, $d$ is the diagonalizable endomorphism in the Jordan decomposition $\overline d$ is its complex conjugate and $L'$ is the derived Lie ALgebra of $L$.
I wanted first to exploit the relation, $p(x) = \overline d$. We get $ad (p(x)) = ad(\overline d). $ Thus, we have the right hand side of (1). We are done if we prove that $p(ad(x)) = ad (p(x)).$ It suffices to provide a counterexample such as $p(x) = x^2,$ showing that the last equality does not hold in general.
Can somebody provide an explicit proof of $(1)$?
Thanks.
Correction: Since $x = d + n$ and given a lemma, according to which, $ad\, x = ad\, d + ad\, n$, the first equality in (1) is straightforward. It only remains to prove the right equality in (1)
