I am working through Volume 1 of Kerf and Bäuerle's book (which I generally find excellent for a humble physicist like myself to learn from), and I'm unfortunately stuck.
In Lemma 11.2.1c, they state that the centre of the KM algebra is contained in the linear span of the simple coroots $\Pi^V$. The proof involves using the explicit realistion $(E, \Pi, \Pi^V)$ of the generalised $n\times n$, rank $r$ Cartan matrix $A$:
$A \equiv \begin{pmatrix}A_r & B\\C&D\end{pmatrix}$ where $A_r$ is a maximal rank $r \times r$ matrix,
$E \equiv \begin{pmatrix}A_r & B&0\\C&D&\mathbb{1}_{n-r}\\0&1_{n-r}&0\end{pmatrix}$ a maximal rank $(2n - r) \times (2n - r)$ matrix.
They consider $c \in Z$ expanded in the basis of $H = \mathbb{C}^{2n -r}$ formed by $\Pi^V \cup\{$ the remaining rows of the matrix realisation $E\}$ (i.e. the basis is just the rows of $E$), use a previous result that states that
$Z = \{h \in H \big| \langle \alpha_i,h\rangle = 0 \;, i = 1,...,n\}$
and end with
[imposing $\langle \alpha_i, c\rangle = 0$ for $i = 1,...n$] one obtains $n$ linearly independent homogeneous equations for the $2 n - r$ coefficients $c_j$. From the fact that $A$ has rank $r$ one easily deduces $c_{n+1} = c_{n+2} = ... = c_{2n - r} = 0$
(emphasis mine). I don't understand why the rank of $A$ tells us this. I've looked at other books and notes and also not found anything illuminating for me.
It's a crucial step for later proving the non-degeneracy of the generalised Cartan Killing form.
Please let me know if I should put more detail, as I'm sure this could be very opaque to someone not intimately familiar with the particular setup (for example, Kac's book does the matrix realisation a little differently).
I think this may be the answer:
Expanding $c$ in the $2n - r$ coefficients $\{c_i\} \cup \{\bar{c}_i\}$, $$c = \sum_i (c_i \alpha_i^V + \bar{c}_i \bar{\alpha}_i^V),$$ where $\alpha_i^V \in \Pi^V$ and $\bar{\alpha}_i^V$ is in the complement of span$\Pi^V$ in $H$, we have
$$ \begin{aligned}\langle \alpha_j,c\rangle &= \sum_i (c_i \langle \alpha_j, \alpha_i^V\rangle + \bar{c}_i \langle \alpha_j, \bar{\alpha}_i^V\rangle)\\ &= \sum_i (A^T_{ji} c_i+ \bar{c}_i \langle \alpha_j, \bar{\alpha}_i^V\rangle)= 0. \end{aligned} $$
Now, by the rank-nullity theorem the matrix $A^T$ has nullity $n-r$ - so there are $n-r$ independent choices of $\{c_i\}$ which will annihilate the first term. For these $n-r$ elements of $Z$, $\bar{c}_i\equiv 0$. But $dimZ = 2n-r - n = n-r$, so this exhausts $Z$.
We know that $\langle \alpha_j, \bar{\alpha}_i^V\rangle \neq 0$ identically because of the non-degeneracy of the Killing form on $H$. Therefore $\bar{c}_i = 0$.
I hope that my logic is sound. Would be interested also to hear if there's a more straightforward path.