# Does an irreducible $M$-matrix have positive diagonal entries?

Does an irreducible $M$-matrix have positive diagonal entries?

An M-matrix $A$ is a matrix which can be expressed as $\alpha I−P$ for some non-negative matrix P and $\alpha\geq \rho(P)$ where $\rho(P)$ is the spectral radius of $P$.

A matrix $A$ is irreducible iff there does not exist a permutation matrix $P$ such that $P^{T}AP==\left( \begin{array}{cc} B & 0\\ C & D \\ \end{array} \right)$. There are many definitions for irreducibility of a matrix. (Reference: Chapter 1 in Nonnegative Matrices in the Mathematical Sciences)

• The structure of $A=\alpha I−P$ suggests that $A$ is irreducible if and only if $P$ is irreducible. Since $P$ is nonnegative and irreducible, I can use Perron-Frobenius Theorem. But I don't know how to use it for this problem. – La Rias Apr 13 '18 at 9:37
As $\alpha I-P=(\alpha+1)\left[I-\frac1{\alpha+1}(I+P)\right]$, we may assume without loss of generality that $\alpha=1$, i.e. $A=I-P$ with $\rho(P)\le1$.
We claim that $A$ has a positive diagonal. Suppose the contrary that $a_{11}\le0$, i.e. $p_{11}\ge1$. Then the first diagonal entry of $P^k$ is also $\ge1$ for every positive integer $k$. Hence $\|P^k\|_\infty\ge1$ and Gelfand's formula implies that $\rho(P)\ge1$. Thus $\rho(P)=1$. As $P$ is irreducible, $v=Pv$ for some positive eigenvector $v$. Therefore $$v_1=\sum_jp_{1j}v_j\ge v_1+\sum_{j>1}p_{1j}v_j,$$ meaning that all off-diagonal entries on the first row of $P$ are zero. This is a contradiction because $P$ is irreducible.