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 $\rho(XAY)\leq \rho(XBY)$.



Characterisation of a matrix ordering property
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 $A\_{ij} \le B\_{ij}$.


