4
$\begingroup$

$A$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Dan Uznanski Aug 21 '14 at 0:18
  • $\begingroup$ 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. $\endgroup$ – M Engineer Aug 21 '14 at 5:00
0
$\begingroup$

There is a simple proof, based on Fiedler's inequality, if your matrix is symmetric.

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 $(A^{-1})_{ii}>1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.