Suppose we have a symmetric matrix $A$ with diagonal elements that are all $0$. Is it necessarily true that $\rho(|A|) \geq \rho(A)$, where $\rho(\cdot)$ is the spectral radius function? Using Gershgorin Circle Theorem, one can only conclude that both $\rho(A)$ and $\rho(|A|)$ lie in the interval $(-\max_i \sum_{j \neq i}a_{ij}, \max_i \sum_{j \neq i}a_{ij})$, but that's where I am stuck. Can anyone help me? Specifically, I am interested to know if there's any truth to the following suppositions:
- $\rho(|A|) < 1 \implies \rho(A) < 1$
- $\rho(A) < 1 \iff \rho(|A|) < 1$
If we equip the space $ \mathbb C^n$ with the euclidian norm $|| \cdot ||_2 $ and define for a matrix $T \in \mathbb C^{n \times n}$ the operator norm by
$$ ||T||= \max\{ ||Tx||_2: ||x||_2=1\},$$
then we have for a symmetric matrix $A$, that
$$\rho(A)=||A||.$$
Now let $A$ be a symmetric matrix. $|A|$ is defined by $|A|=(A^*A)^{1/2}.$ Hence $|A|$ is symmetric.
We get:
$$\rho(|A|)^2=\rho(|A|^2)=\rho(A^*A)=||A^*A||=||A||^2,$$
thus
$$\rho(|A|)=||A||= \rho(A).$$
Remark: this result holds, more general, for symmetic operators on Hilbert spaces.