$\rho(B) \leq \rho(|B|)$, where $\rho$ is the spectral radius

Question

Let $B_{d \times d}$ be an integral matrix and let $A$ be such that $a_{IJ} = |b_{IJ}|$, $1 \leq I, J \leq d$. Then, $\rho(B) \leq \rho(A)$.

My strategy is to use Gelfand's formula, i.e., for a square matrix $M$ and a matrix norm $||\cdot||$, $$\rho(M) = \lim_{k \to \infty} ||M^k||^{\frac{1}{k}}.$$

If I manage to prove that for each pair $1 \leq I, J \leq d$, $(B^n)_{IJ} \leq (A^n)_{IJ}$, the result would follow.

Indeed, for $n=1$ the result is trivially true. Now, suppose that it is valid for $n$. Then, (see problem below the displayed equations)

\begin{align*} (B^{n+1})_{IJ} = (B^{n}B)_{IJ} &= \sum_{K=1}^d (B^{n})_{IK}(B)_{KJ}\\ &\leq \sum_{K=1}^d (A^{n})_{IK}(A)_{KJ}\\ &= (A^{n+1})_{IJ}. \end{align*}

$\textbf{Problem!}$ Isn't that possible that the inequality in the last displayed set of equations does not hold? Because $(B^n)_{IK}$ and $B_{KJ}$ could be negative, for instance, we could have $(A^n)_{IK} = 1, (A)_{KJ} = 1$ and $(B^n)_{IK} = -2, (B)_{KJ} = -1$. My guess is that this cannot happen because $A = |B|$ (entrywise).

Any help is appreciated.

Answer 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

Answer 2

The idea of using Gelfand's formula indeed works. You may prove by mathematical induction that $|(B^n)_{ij}|\le(A^n)_{ij}$.

Answer 3

OP's criterion implies $A^2=\vert B\vert^2 \geq \vert B^2\vert$ (where the absolute value is applied component-wise) by checking triangle inequality on dot products and
$A^k=\vert B\vert^k=\vert B\vert \cdot\vert B\vert^{k-1} \geq \vert B\vert \cdot\vert B^{k-1}\vert\geq \vert B\cdot B^{k-1}\vert=\vert B^{k}\vert$ where the first inequality is an induction hypothesis and the second is from checking triangle inequality on the dot products
$\implies \text{trace}\big(A^k\big)=\text{trace}\big(\vert B\vert ^k\big)\geq \text{trace}\big(\vert B^{k}\vert\big)=\sum_{i=1}^n \vert b^{(k)}_{i,i}\vert\geq \vert\sum_{i=1}^n b^{(k)}_{i,i}\vert =\big \vert\text{trace}\big(B^k\big)\big \vert$
OP's desired result then follows by applying the below lemma

Lemma: $\big \vert\text{trace}\big(B^k\big)\big \vert\leq \big \vert\text{trace}\big(A^k\big)\big \vert$ for all $k\in \mathbb N\cup\big\{0\big\}$ $\implies \rho(B)\leq \rho(A)$
we may assume that $\text{trace}\big(B^k\big)\neq 0$ for some $k\geq 1$, otherwise $B$ is nilpotent and the result trivially holds.

Let $A,B$ have characteristic polynomials $p_A$ and $p_B$. In the annulus $\big\{z: \rho(A) \lt \vert z\vert\lt \infty\big\}$ the rational function $r_A(z) = \frac{p_A'(z)}{p_A(z)}$, has Laurent Series $ = \sum_{k=1}^\infty \text{trace}\big(A^{k-1}\big)\cdot z^{-k}\text{ }$
(check e.g. Cauchy's Integral Formula and $r_A(\infty)=0$)

Equivalently we have analytic $f_A: \Big(0, \frac{1}{\rho(A)}\Big)\longrightarrow \mathbb C$ given by $f_A(z) = r_A\big(z^{-1}\big)= \sum_{k=1}^\infty \text{trace}\big(A^{k-1}\big)\cdot z^{k}$
where $0$ is a removable singularity that we've defined over. I.e. $f_A$ converges absolutely for $\vert z \vert \lt \frac{1}{\rho(A)}$ and has a pole at some $\vert z \vert = \frac{1}{\rho(A)}$. Thus for $z\in \Big(0, \frac{1}{\rho(A)}\Big)$ we have

$\big \vert \sum_{k=1}^\infty \text{trace}\big(B^{k-1}\big)\cdot z^{k}\big \vert \leq \sum_{k=1}^\infty \big \vert\text{trace}\big(B^{k-1}\big)\big \vert\cdot \big \vert z \big \vert^{k}\leq \sum_{k=1}^\infty \big \vert\text{trace}\big(A^{k-1}\big)\big \vert\cdot \big \vert z \big \vert^{k}\lt \infty$
so $f_B$ converges absolutely in $\Big(0,\frac{1}{\rho(A)}\Big)\implies \frac{1}{\rho(B)}\geq \frac{1}{\rho(A)}\implies \rho(B)\leq \rho(A)$

remarks:

• $A$ and $B$ both being the same size isn't needed though it simplifies the presentation
• The above lemma also applies when $A\geq B \geq \mathbf 0$ which leads to a fairly easy inductive proof of Perron Theory in the case of positive $A$