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Can we say that there is a relationship between the eigenvalues of the matrix $A$ and its absolute value as $B$, where $b_{ij}=|a_{ij}|$? Consequently, can we say that there is a relationship between the spectral radius of them?

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    $\begingroup$ Almost surely, no relationship at all. $\endgroup$ – Jean Marie Dec 8 '17 at 7:02
  • $\begingroup$ Agree with Jean that there's no chance of a nice relationship in general. If $A$ is diagonally dominant, though, the spectral radii will be similar, since the two matrices have the same Gershgorin disk radii. $\endgroup$ – user7530 Dec 8 '17 at 7:17
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Let $A\in M_n(\mathbb{C})$. According to Wielandt, if $B$ is irreducible, then $\rho(A)\leq \rho(B)$.

EDIT. In fact, the inequality remains true even if $B$ is reducible.

Indeed, there is a sequence of positive matrices $(B_k)$ s.t. $B\leq B_k$ and $(B_k)$ tends to $B$.

Thus $|A|\leq B_k$ and $B_k>0$ implies that $\rho(A)\leq \rho(B_k)$. The conclusion comes from $\rho(B_k)$ tends to $\rho(B)$.

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There are special cases when the two spectra are closely related. E.g., assume you can write

$$|A| = \Lambda A\Lambda$$

where $\Lambda$ is a diagonal matrix with entries $\pm 1$. Since this a similarity transformation, $A$ and $|A|$ have the same eigenvalues with same multiplicity. There are special cases when this is possible, e.g.,

$$A=\begin{pmatrix}a & -b \\ -c & d\end{pmatrix}$$

Choose

$$\Lambda = \begin{pmatrix}1 & 0 \\ 0 & -1\end{pmatrix}$$

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