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Let $n$ be a positive integer; we consider all matrices mentioned henceforth to be $n$-by-$n$ matrices. Let $A$ and $B$ be matrices wherein all entries are nonnegative (such matrices will be called nonnegative matrices). Denote by $\rho(X)$ the spectral radius of a matrix $X$, i.e. the modulus of the largest eigenvalue of $X$.

Then if $A\leq B$ (i.e. $A_{ij}\leq B_{ij}$ for all $i,j$), we have that for any nonnegative matrices $X$ and $Y$, $\rho(XAY)\leq \rho(XBY)$.**

Is the converse true? That is, if $\rho(XAY)\leq \rho(XBY)$ for any nonnegative matrices $X$ and $Y$, does this imply that $A\leq B$? If not, can you find a necessary and sufficient condition on $A$ and $B$ for this condition to hold? Sufficient conditions which are weaker than $A\leq B$ will also be appreciated.

** $XAY\leq XBY$ and $XAY$ and $XBY$ are both nonnegative matrices; from this it can be shown that $\rho(XAY)\leq \rho(XBY)$.

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If we take $X = E_{1,i}$ (the matrix with $X_{1,i} = 1$ and all other entries $0$) and $Y = E_{j,1}$, $XAY$ has $(1,1)$ entry $A_{ij}$ and all others $0$, and its spectral radius is $A_{ij}$. So $\rho(XAY) \le \rho(XBY)$ says $A_{ij} \le B_{ij}$.

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