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

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?

• What do you mean by the absolute value of the matrix? Elementwise absolute value? – Robert Israel Jan 16 '18 at 20:49
• Yes, the absolute value of the matrix is exactly the elementwise absolute value. – M. Raha Jan 17 '18 at 13:57

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 $7$.
• 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. – Robert Israel Jan 18 '18 at 21:51