# Matrices whose inverse is positive

I have recently come across some examples of matrices with a special structure. I will describe these matrices here and I hope that somebody will be able to point out a source where I can find more information about them. Consider an $$n\times n$$ matrix $$A$$ with elements $$a_{ij}$$ having the following properties. The elements with $$i=j$$ (call them $$b_i$$) are negative. The elements with $$j=i+1 {\rm mod} n$$ (call then $$c_i$$) are positive. All other elements are zero. The determinant of a matrix of this type is $$\prod_i b_i+(-1)^{n+1}\prod_i c_i$$. A property of these matrices which I found surprising is that $$(-1)^{n+1}(\det A)A^{-1}$$ is a positive matrix, i.e. all its entries are positive. I found this by playing around with some examples. Can anybody point out to me some general theory which explains this observation? I met these matrices repeatedly when looking at certain chemical reaction networks. In that context the positivity statement is valuable because it allows the Perron-Frobenius theorem to be applied.

A class of matrices with entrywise positive inverses (inverse-positive matrices) appears in a variety of applications and has been studied by many authors. See, for example,

M-Matrices Whose Inverses Are Totally Positive

or

Positive, path product, and inverse M-matrices

In these papers (and those referred to by them) you will find methods to construct other classes of matrices with this property.

• It seems that these papers deal with $M$-matrices, which, it seems, means matrices $A$ such that: (i) $A$ is invertible and every entry of $A^{-1}$ is nonnegative; and (ii) The off-diagonal entries of $A$ are nonpositive. By contrast, the question seems to refer to matrices which satisfy (i) but for which the off-diagonal entries are positive (plus other conditions). Is there a clear way to relate the question to $M$-matrices? – Aaron Tikuisis Mar 18 '12 at 10:52
• I support Aaaron's comment. Read about $M$-matrices! – Denis Serre Mar 18 '12 at 13:21
• The possibility to change some signs follows from David Speyer's answer. – Anatoly Kochubei Mar 18 '12 at 19:07
• Maybe I'm a little slow, but I still don't see it. David seems to have simply computed the inverse and shown that the entries are nonnegative - which does answer the question, but doesn't seem to shed light on where $M$-matrices come into play. – Aaron Tikuisis Mar 19 '12 at 5:47

This is straightforward from the adjoint formula for the inverse matrix. Let $A_{ij}$ be the matrix formed by deleting row $i$ and column $j$ from $A$. We must show that $(-1)^{n+1} (-1)^{i-j} \det A_{ij} > 0$.

We can reorder the rows and columns of $A$ cyclically to assume without loss of generality that $j=n$. Then $A_{in}$ is block diagonal with two blocks of size $i-1$ and $n-i$. The first block is upper diagonal with diagonal entries $b_1 b_2 \cdots b_{i-1}$; the second block is lower diagonal with diagonal entries $c_{i+1} c_{i+2} \cdots c_n$. So $\det A_{ij}$ has sign $(-1)^{i-1} = (-1)^{n+1} (-1)^{n-i}$ as desired.

In particular, we have an explicit formula for $A^{-1}$. The entry $(A^{-1})_{ij}$ is $$(-1)^{i-j} \frac{b_{j+1} b_{j+2} \cdots b_{i-1} c_{i+1} c_{i+2} \cdots c_j}{\det A}.$$

Matrices whose inverses are nonnegative are also called monotone. There are a number of equivalent characterizations in Theorem 6.2.3 of the wonderful book by Berman and Plemmons:

In the case of this question, the matrix might not be an $M$-matrix. It depends on the actual entries.