Let $\mathfrak{g}$ be a simple finite-dimensional Lie algebra. Let $\mathfrak{h}$ be a Cartan subalgebra, $C$ the Cartan matrix, and $R$ a system of simple roots $\left(\alpha_{1},\cdots,\alpha_{n}\right)$.
Using the following fact
The Killing form $Tr\left(ad_{X}\circ ad_{Y}\right)$ induces a bilinear symmetric non-degenerate form on $\mathfrak{h}^{*}$
how can it be shown that there exists a symmetric matrix $A$ such that $$ A=diag\left(r_{1},\cdots,r_{n}\right)C $$ where $r_{j}$ are prime and strictly positive $\forall j$?
Note: This question is related to exam prep for an advanced graduate course "Lie Theory and Representations 2". So far, I figured that if $\mathfrak{h}$ is considered in diagonalized form, $ad$ can be represented by a $diag$ matrix, and I thought about applying Serre relations $$ \left(ad_{e_{i}}\right)^{1-C_{i,j}}\left(e_{j}\right)=0\forall i\neq j $$ and $$ \left(ad_{f_{j}}\right)^{1-C_{i,j}}\left(f_{j}\right)=0\forall i\neq j $$ to get the $C$ part, but so far, I could not get it to stick. I think I may be on the right path?
Am I missing something here? If I take $r_i=(\alpha _i, \alpha _i)$, and $D$ the diagonal matrix with entries $r_i$, then clearly $A=DC$ is symmetric, where $C$ is the Cartan matrix.
Now, if $\mathfrak g$ is a simple Lie algebra, then the numbers $r_i$ are either all $1$, or exactly one of them is $2$, a prime, or exactly one of them is $3$ (a prime). The condition that the $r_i$ be prime, seems strange in this context.