Are the entries of the cholesky decomposition of a symmetric positive definite inverse M-matrix nonnegative?

Let $\Sigma$ be a symmetric positive definite matrix such that $\Sigma^{-1}$ is a $M$-matrix.

(A $n\times n$ matrix $T$ is called a $M$-matrix of it is of the form $T = \lambda \mathrm{I}_n - P$, where $P$ has nonnegative elements and $\lambda > 0$ exceeds the absolute value of every characteristic root of $P$.)

Assume that we have the Cholesky decomposition $\Sigma = L L^{\top}$.

Are the entries of $L$ nonnegative ?

Note : If we assume that $\Sigma$ is a symmetric positive definite matrix with nonnegative entries, then there are easy counterexamples. Hovever, asking that $\Sigma^{-1}$ is a $M$-matrix is a stronger condition than asking that the entries of $\Sigma$ be nonnegative, so I'm not sure.

Solution 1:

Yes. In fact, one can prove the following stronger statement, which is modified from theorem 2.3 in chapter 6 of Nonnegative Matrices in the Mathematical Sciences by Berman and Plemmons:

Let $M$ be an $M$-matrix that is not necessarily symmetric. If all trailing principal minors of $M$ are positive, then $M^{-1}$ admits an LU decomposition where the $L$ and $U$ are nonnegative.

This can be proved by mathematical induction on the size of $M$. In the inductive step, we may assume that $$ M=\pmatrix{a&x^T\\ y&(L_0U_0)^{-1}} $$ where $x,y\le0$ because $M$ is an $M$-matrix and (by induction assumption) $L_0,U_0$ are respectively a nonnegative lower triangular matrix and a nonnegative upper triangular matrix. Since all trailing principal minors of $M$ are positive, the Schur complement $b:=a-x^TL_0U_0y$ is positive. Thus $$ M=\pmatrix{a&x^T\\ y&(L_0U_0)^{-1}} =\pmatrix{1&x^TL_0\\ 0&U_0^{-1}}\pmatrix{b&0\\ U_0y&L_0^{-1}} =:U^{-1}L^{-1} $$ where $$ L=\pmatrix{b&0\\ U_0y&L_0^{-1}}^{-1} =\pmatrix{b^{-1}&0\\ -L_0U_0yb^{-1}&L_0} $$ and $$ U=\pmatrix{1&x^TL_0\\ 0&U_0^{-1}}^{-1} =\pmatrix{1&-x^TL_0U_0\\ 0&U_0} $$ are nonnegative. In your case, put $M=\Sigma^{-1}$.