# When does a strictly diagonally dominant matrix have dominant principal minors?

$A$ is an $N\times N$ matrix with diagonal elements $a_{ii}=1-s_{i}$, and off diagonal elements $a_{ij}=s_{i}w_{ij}$ for $i≠j$. Assume $0≤s_{i}<1/2$ and $\sum_{j≠i}w_{ij}=1$ for all $i$ and $0≤w_{ij}≤1$. As $1-s_{i}>∑_{j≠i}a_{ij}=s_{i}∑_{j≠i}w_{ij}$ for all $i$, matrix $A$ is strictly row diagonally dominant. Thus, $\det(A)>0$ and has positive principal minors, $M_{ii}$ where $ii$ is the submatrix from eliminating row $i$ and column $i$. (see Tsatsomeros 2002). There exists an inverse matrix $B=A^{-1}$.

I want to show that the diagonal elements $b_{ii}=\frac{M_{ii}}{\det(A)} > 1$. I am able to show this for $N=3$ and $4$. For example, by expanding across the top gives $b_{11}>1$ as $M_{11}>-ω_{12}M_{12}+ω_{13}M_{13}-ω_{14}M_{14}$ , which follows from $M_{11}>‖M_{1j}‖_{j≠1}$. Here the results follows if the principal minor is dominant in the sense of being larger than the minors along the same row.

I do not know how to extend this to $N>4$ and have not found the result in my imperfect search of the literature.

• I'm not fully up on the literature but I included a link to "On the Cayley Transform of Positivity Classes of Matrices" math.technion.ac.il/iic/ela/ela-articles/articles/… because it looked like the one you meant. – Dan Uznanski Aug 21 '14 at 0:18
• The A matrix is non-negative. The article refers to the result for matrices with non-positive off diagonals, an M-matrix. However, A=I-S , where S has on the diagonal $s_{i}≥0$ and on the off diagonal $-sw_{ij}≤0$. Thus, S may be an M-matrix, but I don't see how the result follows for A. – M Engineer Aug 21 '14 at 5:00

If A is symmetric then A is positive definite. By Fiedler's inequality $A\circ A^{-1}-Id$ is positive semidefinite, where $A\circ A^{-1}$ stands for the Hadamard product of $A$ by $A^{-1}$.
Since $A_{ii}=1-s_i<1$ and $A_{ii}(A^{-1})_{ii}-1\geq 0$, because $A\circ A^{-1}-Id$ is positive semidefinite, then $(A^{-1})_{ii}>1$.