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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 $\sum_i b_i+(-1)^{n+1}\sum_i c_i$. 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.

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This is not important, but shouldn't be \prod_i instead of \sum_i? – Cristi Stoica Mar 18 '12 at 11:25
Yes it should be \prod_i. Sorry about that carelessness. – hydrobates Mar 18 '12 at 12:13
up vote 6 down vote accepted

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,


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

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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 $$ (-1)^{i-j} \frac{b_{j+1} b_{j+2} \cdots b_{i-1} c_{i+1} c_{i+2} \cdots c_j}{\det A}.$$

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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.

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