Largest eigenvalue of entry-wise absolute value of a matrix w.r.t. original maximum eigenvalue [duplicate]

Solution 1:

start by assuming WLOG that $A$ is strictly positive.
i.e. if needed consider $B':= \delta\mathbf {11}^T + B$ and $A':= \delta\mathbf {11}^T + A$ for any $\delta \gt 0$ and topological continuity of eigenvalues gives the results.

main idea
Perron theory tells us there is some positive vector $\mathbf v$ such that
$A\mathbf v = \lambda_1 \mathbf v$ for $\lambda_1 \gt 0$

since similarity transforms preserve eigenvalues, use a well chosen similarity transform
$D := \text{diag}\big(\mathbf v\big)$

then consider
$\big(D^{-1}AD\big)\mathbf 1=D^{-1}A\big(D\mathbf 1\big) = D^{-1}A\mathbf v = \lambda_1 D^{-1}\mathbf v = \lambda_1 \mathbf 1$
so the one's vector is the Perron vector, and $\big(D^{-1}AD\big)$ has homogeneous row sums. In the case of $\lambda_1=1$ this would be called a stochastic matrix.

application of Gerschgorin discs tells us that
$\lambda_\text{max modulus}\big(D^{-1}BD\big)\leq \lambda_1\longrightarrow \lambda_\text{max modulus}\big(B\big) \leq \lambda_\text{max modulus}\big(A\big) = \lambda_1$
and completes the exercise