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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. For more information see this paper: http://www.math.technion.ac.il/iic/ela/ela-articles/articles/vol20_pp90-94.pdf |
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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. For more information see this paper: http://www.math.technion.ac.il/iic/ela/ela-articles/articles/vol20_pp90-94.pdf |
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