There is a theorem that any finite dimensional semisimple Lie algebra $L$ is in fact a direct sum of simple ideals $I_1, I_2, …, I_n$, and I am now studying its proof in Humphreys' "Introduction to Lie algebras" (section 5.2).
It starts with any ideal $I$ in $L$. We introduce $$I^{\bot}=\{x \in L: \kappa(x,y) = \mathrm{Trace}(ad_x ad_y) = 0, \forall y\in L\},$$
which is an ideal, and by the Cartan's Criterion and semi-simplicity of L, we have $I \cap I^{\bot} = 0$, which I understand.$\\$
The fact I need is that $I \oplus I^{\bot} = L$, or, equivalently $\mathrm{dim} \ I + \mathrm{dim} \ I^{\bot} = \mathrm{dim} \ L$. $\\$
In other source I found that it is implied by the non-degenarativity of a Killing form $\kappa(x,y) = \mathrm{Trace}(ad_x ad_y)$ for a semisimple $L$
$\left( \text{ non-degenerativity means that the set } S = \{x \in L: \kappa (x,y) = 0, \ \forall \ y \in L \} \text{ is equal to 0} \right)$, but I don't see how.
Best Answer
The fact that the bilinear form $b$ defined on $L$ is non degenerated is equivalent to the fact that $d(b):L\rightarrow L^*$ defined by $d(b)(x)(y)=b(x,y)$ is an isomorphism. $I^{\perp}$ is identified with the anihilator $I^0$ of $I$ with this isomorphism and $dim(I)+dim(I^0)=dim(L)$.