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Let a matrix $A\in \mathbb{R}^{n\times n}$ is SDD(strictly diagonally dominant). How we can show $\rho(A)=\rho(|A|)$ ($\rho(A)$ represent the spectral radius of the matrix $A$ and $|A|$ represent the absolute value of the matrix $A$). By using Gershgorin's theorem can can we say that $\rho(A)=\rho(|A|)$ for the the non SDD matrix $A$ with positive diagonal entries?

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  • $\begingroup$ What do you mean by the absolute value of the matrix? Elementwise absolute value? $\endgroup$ – Robert Israel Jan 16 '18 at 20:49
  • $\begingroup$ Yes, the absolute value of the matrix is exactly the elementwise absolute value. $\endgroup$ – M. Raha Jan 17 '18 at 13:57
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It's not true. For example, the SDD matrix $$ A = \pmatrix{5 & 1 & 1\cr 1 & 5 & 1\cr 1 & -1 & 5\cr}$$ has spectral radius $6$, while $|A|$ has spectral radius $7$.

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  • $\begingroup$ This is nice! How did you construct it? $\endgroup$ – user1551 Jan 18 '18 at 10:57
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    $\begingroup$ Well, I know that the spectral radius of a matrix with positive entries and all row sums $=k$ is $k$, so I took one of those and changed one sign. The fact that the eigenvalues turned out to still be integers is a bit of luck. $\endgroup$ – Robert Israel Jan 18 '18 at 21:51
  • $\begingroup$ Thanks for sharing your thought process. $\endgroup$ – user1551 Jan 18 '18 at 22:26
  • $\begingroup$ Thanks for helping to the more clarity the problem by the interesting counterexample. $\endgroup$ – M. Raha Jan 19 '18 at 8:09

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