About some details about the proof that real Lie algebra with positive Killing form is zero

In the post about proving that real Lie algebra with positive Killing form is zero: real Lie algebra with positive Killing form is zero:

Let $L$ be a real Lie algebra with positive definite Killing form. Its Killing form $\kappa$ defines an inner product on $L$. Hence $L$ is reductive. Thus the quotient $L/Z(L)$ is semisimple. So, the Killing form is negative definite of $L/Z(L)$. Therefore, this Killing form is both positive definite and negative definite, it follows that $L/Z(L) = {0}$. So we get $L = Z(L)=\ker(\kappa)$. But $\kappa$ is non-degenerate since it’s positive definite. It follows that $L= {0}$.

I am confused with the following gaps:

1. Why killing form on $L/Z(L)$ is negative definite?
2. Why the induced Killing form on $L/Z(L)$ is positive definite?

And is there any relation with the fact that $\mathfrak{g}$ is real?

Here is another argument: if $\mathfrak{g}$ has positive-definite Killing form $\kappa$, let $x_1,...,x_n$ be an orthonormal basis of $\mathfrak{g}$ and write $[x_i,x_j] = \sum_k a_{ijk} x_k$ with $a_{ijk} \in \mathbb{R}$. The invariance $\kappa([x,y],z) = \kappa(x,[y,z])$ implies $a_{ijk} + a_{ikj} = 0$. Therefore $$\kappa(x_i,x_i) = \mathrm{tr}( \mathrm{ad}(x_i) \cdot \mathrm{ad}(x_i)) = \sum_{j,k} a_{ijk}a_{ikj} = - \sum_{j,k} a_{ijk}^2 < 0,$$ contradiction.
• $a_{ijk} = K([x_{i},x_{j}],x_{k}) = K(x_{i}, [x_{j}, x_{k}]) = -K([x_{i}, x_{k}], x_{j}) = -a_{ikj}$ – Rob Rockwood Jan 27 at 12:24