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I have a nonnegative matrix which is sort of diagonally dominant but not quite.

I have $a_{ii}>a_{ij}$ and $a_{ii}>a_{ji}$ for any $i$ and $j$ not equal to $i$.

I wonder whether it must be nonsingular.

lioqq
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1 Answers1

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Not when $A$ is at least $3\times3$. E.g. $\pmatrix{5&0&3\\ 0&5&4\\ 3&4&5}$ is singular.

More generally, pick any unit vector $u$ such that $0<u_i<1$ for each $i$. Pick also some $\epsilon\ge0$. The matrix $A$ below is then a singular nonnegative matrix: $$ \underbrace{\left[\pmatrix{I&u\\ u^T&1}+\epsilon\pmatrix{uu^T&u\\ u^T&1}\right]}_A\pmatrix{u\\ -1}=0. $$ ($A$ is entrywise positive if $\epsilon>0$.) When $\epsilon\ge0$ is sufficiently small, we have $a_{ii}>\max(a_{ij},a_{ji})$ for all $i\ne j$.

user1551
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  • Thanks. That's very helpful. What if I add the condition that each row is a probability vector? – lioqq Dec 04 '20 at 04:45