Let $A \in \operatorname{GL}_n(\mathbb R)$ be a symmetric matrix which is positive definite, i.e. $A = Q^tQ$ for some invertible matrix $Q$. Suppose that the off diagonal entries of $A$ are $\leq 0$. Must all the entries of $A^{-1}$ be positive?
I am asking because the Cartan matrix $A = (\langle \alpha_i, \alpha_j^{\vee} \rangle)$ of any reduced root system with base $\alpha_1, ... , \alpha_l$ is, after factoring out positive scalars out of each column, a symmetric positive definite matrix with this property. The inverse of this matrix gives a formula for how to write the fundamental weights $\omega_i$ (that is, the dual basis to the simple coroots) as linear combinations of simple roots. I want to say that each fundamental weight is a positive linear combination of simple roots, which is equivalent to the assertion that $A^{-1}$ has all positive entries.
Yes. If you partition $A$ as $\pmatrix{a&b^T\\ b&D}$, then $A^{-1}=\pmatrix{s^{-1}&-s^{-1}b^TD^{-1}\\ \ast&\ast}$, where $s=a-b^TD^{-1}b$ is the Schur complement of $D$ in $A$. Since $A$ is positive definite, $s^{-1}$ must be positive. Therefore, by mathematical induction, $D^{-1}$ and in turn $-s^{-1}b^TD^{-1}$ are also entrywise positive. Hence all entries on the first row of $A^{-1}$ are positive. Similarly, the entries on other rows are positive too.