In what cases can the spectral radius of a matrix and the spectral radius of its absolute value be equated?

Question

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?

Answer 1

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 $7$.