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Well, $B$ is the Hadamard product of $A$ with itself, so if $A$ is nonnegative or psd, its radius is bounded from above by the square of the radius of $A$. For real matrices, at least.

Well, $B$ is the Hadamard product of $A$ with itself, so its radius is bounded from above by the square of the radius of $A$. For real matrices, at least.