Let $G$ be a linear algebraic group whose Lie algebra $\mathfrak{g}$ is semisimple. Let $x$ be a regular semisimple element of $G$. Write $\mathfrak{t}$ for the Lie algebra of the maximal torus $T = C(x)$ centralizing $x$, and let $\mathfrak{h}$ be the image of $\mathfrak{g}$ under the map $g\mapsto \textrm{Ad}_x(g)-g$.
Since $g$ is semisimple, we may diagonalize it; then all the elements of $\mathfrak{h}$ have only zeroes in the diagonal. It is not hard to see that, for $G$ a classical group ($\textrm{SL}_n$, $\textrm{SO}_n$, $\textrm{Sp}_{2 n}$, over an arbitrary field), the spaces $\mathfrak{t}$ and $\mathfrak{h}$ are orthogonal under the Killing form: the Killing form is then a multiple of $(X,Y)\mapsto \textrm{tr}(X Y)$, and, for $X$ diagonal and $Y$ having only zeroes in the diagonal, $X Y$ clearly has only zeroes in the diagonal, and thus has trace $0$.
Is there a clean proof that works for all linear algebraic groups $G$ with $\mathfrak{g}$ semisimple -- preferably one with as little computation or casework as possible?
