Consider a square matrix $A=DS$ where $S$ is symmetric with diagonal entries being $0$ and $D$ is a diagonal matrix for normalizing $S$'s row sums so that $Ae=e$ where $e$ is a vector with all entries being $1$ assuming none of the row sums of $S$ is $0$. Also, $A$ has at least one negative entry. Can anyone prove/disprove that the spectral radius of $A$, $\rho(A)$, satisfies $\rho(A)>1$? Thank you.

UPDATE: I am very sorry but I've left out an important condition that "$A$ has at least one negative entry".


Consider $S=\pmatrix{0&-1&0\\-1&0&2\\ 0&2&0}$. Then $A=\pmatrix{0&1&0\\-1&0&2\\ 0&1&0}$ and the eigenvalues of $A$ are $-1,0,1$. Hence $\rho(A)=1$.

  • $\begingroup$ I am sorry but I have left out a condition: at least one entry of $A$ is negative. $\endgroup$
    – ziyuang
    Apr 17 '13 at 16:23

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