Producing a nontrivial proper ideal in the (Kac-Moody) Lie algebra $\mathfrak{g}(A)$

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Proposition 1.7 in Kac's book Infinite dimensional Lie algebras states that if $A=(a_{ij})$ is any complex matrix (that is, not necessarily a generalized Cartan matrix) then the Lie algebra $\mathfrak{g}(A)$ is simple if and only if $\det(A)\neq 0$ and for every pair of indices $i,j$ the following condition holds:

$(S_{ij})$ There exist indices $i_1,\dots,i_s$ such that $a_{ii_1}a_{i_1i_2}\cdots a_{i_{s-1}i_s}a_{i_sj}\neq 0$.

He gives no proof of the necessity of the conditions saying that they are obviolsy necessary.

Well, it is really easy to see that if $\mathfrak{g}(A)$ is simple, then $\det(A)\neq 0$. Now I want to prove that condition $(S_{ij})$ is satisfied for all indices $i,j$ and my idea is to prove that if this condition does not hold then I can produce a non-trivial proper ideal of $\mathfrak{g}(A)$. So I let $$ I = \{i \; | \; \text{for all } j \text{ condition } (S_{ij}) \text{ is satisfied}\}. $$ By assumption, $I\neq \{1,\dots,n\}$. Then I would like to take some ideal attached to the set $I$, but I cannot think of one. Maybe taking something as the ideal $\mathfrak{a}$ generated by the Chevalley generators $e_i$ and $f_i$ for $i\in I$ (so that in particular $\alpha_i^\vee\in \mathfrak{a}$ for all $i\in I$), but I cannot prove that this ideal is proper.

If $A$ satisfies the additional condition that $a_{ij}=0$ implies $a_{ji}=0$, then I can prove that $\mathfrak{g}(A)$ decomposes as a direct sum of non-zero ideals, but this extra hypothesis is not assumed in Proposition 1.7.

Any suggestions?