The leading-block and eigenvalue conjecture for E0_2-matrices

In fact, the ENTIRE inverse is a $\mathbf Z$-matrix, which strengthens the assertion concerning only its leading principal block.

Let $A\in\mathbf Z\cap\mathbf E_{0_2}$. Theorem 3.9 gives

$$\det A[S,S]\ge0 \quad (|S|\le n-2),$$ $$\det A[S,S]<0 \quad (|S|=n-1), \qquad \det A<0.$$

Set $\det A[\varnothing,\varnothing]=1$.

For distinct indices $i,j$, expand the off-diagonal cofactor according to its simple directed $i$-to-$j$ path:

$$\operatorname{adj}(A)_{ij} = \sum_{P:i\rightsquigarrow j} (-1)^{\ell(P)} \left(\prod_{(u,v)\in P}a_{uv}\right) \det A[V\setminus V(P),V\setminus V(P)].$$

Indeed, each term in the cofactor permutation expansion decomposes uniquely into its directed $i$-to-$j$ path and a permutation on the complementary vertices.

Because $A$ is a $\mathbf Z$-matrix, every off-diagonal edge weight satisfies $a_{uv}\le0$. Consequently

$$(-1)^{\ell(P)}\prod_{(u,v)\in P}a_{uv}\ge0.$$

Every such path contains at least two vertices, so its complementary principal minor has order at most $n-2$ and is nonnegative. Every term in the path expansion is therefore nonnegative:

$$\operatorname{adj}(A)_{ij}\ge0\qquad(i\ne j).$$

Since $\det A<0$,

$$(A^{-1})_{ij} =\frac{\operatorname{adj}(A)_{ij}}{\det A}\le0 \qquad(i\ne j).$$

Thus $A^{-1}$ itself is a $\mathbf Z$-matrix, and hence so is each of its principal diagonal blocks, including the specified leading block.

For the eigenvalue assertion, write

$$\det(tI+A) =t^n+c_1t^{n-1}+\cdots+c_{n-2}t^2+c_{n-1}t+c_n, \qquad c_k=\sum_{|S|=k}\det A[S,S].$$

The established principal-minor signs imply

$$c_k\ge0\quad(1\le k\le n-2),\qquad c_{n-1}<0,\qquad c_n<0.$$

After zero coefficients are omitted, this coefficient sequence has exactly one sign change. Descartes' rule of signs therefore gives exactly one positive real root of $\det(tI+A)$, counted with multiplicity. Equivalently, $A$ has exactly one negative real eigenvalue. Both conjectured assertions follow.