Orthonormal basis of Cartan subalgebra relative to Killing form

I'm trying to understand a step in a proof:

Let $\mathfrak{g}$ be semi-simple (finite dimensional) Lie-algebra over $\mathbb{C}$, $\mathfrak{h}\subset\mathfrak{g}$ a Cartan subalgebra and let $\kappa:\mathfrak{g}\times\mathfrak{g}\to\mathbb{C}$ be the Killing form.

In this setting, the author of the proof chooses an orthonormal basis $h_1,\dots,h_n$ of $\mathfrak{h}$ relative to the Killing form, which is - to my understanding - a basis satisfying $\kappa(h_i,h_j)=\delta_{ij}$.

Why is it always possible to find such an orthonormal basis?

Thank you for your help!

• Thus the algebraic closure of $\mathbb{C}$ is absolutely crucial to the classification of (semi-?)simple complex Lie Algebras. – JP McCarthy Jan 16 '13 at 15:12
• If you have an orthonormal basis on a complex vector space, aren't there always nonzero vectors which are orthogonal to themselves? $$(ie_1 + e_2,ie_1 + e_2) = -1 + 1 = 0$$ – D_S Dec 24 '17 at 5:25
• But the Killing form is not a complex inner product, it is a symmetric bilinear form over $\mathbb{C}$. – D_S Dec 24 '17 at 13:22
The Killing form is symmetric and non-degenerate(Cartan's criterion). For such bilinear forms you can always diagonalize it via a proper basis. So in particular over $\mathbb{C}$ you should be able to find an orthonormal basis.