$V$ vector space over $\mathbb{C}$, $L$ Lie subalgebra (subspace and closed under Lie bracket) of $gl(V)$, linear maps $V \to V$. Suppose $d \in L$ diagonalisable, show that $\overline{ad(d)} = ad(\bar{d})$ (bar represents complex conjugate).
I have that $d$ is diagonalisable so there exists a basis $B$ of $V$ such that $[d]_B$ is diagonal, where $[d]_B$ is the matrix of $d$ with respect to $B$.
I think I am missing something obvious, we need to show that $\forall x \in L, \overline{ad(d)}(x) = \overline{[d, x]} = \overline{dx - xd}$ is equal to $\bar{d}x - x\bar{d} = [\bar{d}, x] = ad(\bar{d})$ but I just can't think of how to show $\overline{dx - xd} = \bar{d}x - x\bar{d}$.
I thought we could work with matrices but I run into the same problem, I'm sure I'm missing something obvious.
Any hints or help appreciated, thanks.
Here is a counterexample. Let $L=\mathfrak{gl}_2(\mathbb{C})$ and $d=\begin{pmatrix} 1 & 0 \cr 0 & i \end{pmatrix}$, $x=\begin{pmatrix} 0 & i \cr 0 & 0 \end{pmatrix}$. Then we have $$ \overline{[d,x]}=\overline{dx-xd}=\begin{pmatrix} 0 & 1-i \cr 0 & 0 \end{pmatrix}, $$ but $$ [\overline{d},x]=\overline{d}x-x\overline{d}=\begin{pmatrix} 0 & i-1 \cr 0 & 0 \end{pmatrix}. $$ Do you have a link for the question ?