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$V$ 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 $bar{d}x - xbar{d} = [bar{d}, x] = ad(bar{d})$ but I just can't think of how to show $overline{dx - xd} = bar{d}x - xbar{d}$.

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
$$
[overline{d},x]=overline{d}x-xoverline{d}=begin{pmatrix} 0 & i-1 cr 0 & 0 end{pmatrix}.
$$
Do you have a link for the question ?

The equality you want to prove is $overline{ad(d)} = adoverline{(d)}$ which is not equivalent to $overline{dx - xd} = bar{d}x - xbar{d}$.

If $d = text{diag}(lambda_1, dots, lambda_n)$, then $ad(d) = text{diag}(lambda_i - lambda_j)$, where $1 leq i,j leq n$. More precisely, $ad(d)$ has eigenvalues $lambda_i - lambda_j$ corresponding to eigenvectors $E_{ij}$. Therefore $$overline{ad(d)} = overline{text{diag}(lambda_i - lambda_j)} = text{diag}(overline{lambda_i - lambda_j}) = text{diag}(overline{lambda_i} - overline{lambda_j}) = ad(overline{d}).$$