I am currently studying Lie algebras and stumbled upon this little lemma in my lecture notes. The proof given there is insufficient I believe. The proof there is based on choosing an "orthonormal" basis $\{T_1, \ldots, T_n\}$ where $T_1 = \frac{x}{\sqrt{\kappa \left(x, x\right)}}$ and $\kappa (T_a, T_b) = \delta_{ab}$. Then what's left to show is that $ad(T_1)$ is diagonalizable and then in the lecture notes it is argued that because $ad(T_1)$ is basically just given by structure coefficients in the given basis, it is antisymmetric and thus diagonalizable. But not every antisymmetric matrix is diagonalizable, this would only be true if it was a real-valued matrix over $\mathbb{C}$, but it is not. It is complex valued. So my question here is, is this proof valid and am I thus just not getting something here, or is it actually insufficient? And does anyone have an alternative/actual proof?
Thanks in advance! :)
If by fssc you mean "finite dimensionnal semisimple complex", I think that $x=\begin{pmatrix}-2&0&0\\0&1&1\\0&0&1\end{pmatrix}\in\mathfrak{sl}_3$ satisfies $\kappa(x,x)\neq0$ and $ad_x$ is not diagonalizable.
Diagonalizable is equivalent to semisimple and $ad_x$ is semisimple iif $x$ is semisimple (as an element of $\mathfrak{gl}(V)$). So $ad_x$ is diagonalizable iif $x$ is diagonalizable. The Killing form is a non zero multiple of $(x,y)\mapsto Tr(xy)$ so $\kappa(x,x)$ is non-zero iif the sum of the square of the eigenvalues of $x$ is non-zero. The two properties seems relatively uncorelated.
You can check Chapter 20 of Lie algebras and algebraic groups by P.Tauvel and R.Yu for the properties aforementionned.