Entrywise modulus matrix and the largest eigenvector

Question

Disclaimer. This is a cross-post from math.SE where I asked a variant of this question two days ago which has been positively received but not has not received any answers.

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Let $A$ be a complex Hermitian $n\times n$ matrix and define the matrix $B$ to be the entrywise absolute value of $A$, i.e., $B_{ab}=\lvert A_{ab}\rvert$. Furthermore suppose that $B$ has only one eigenvalue of multiplicity one with maximal modulus $\lvert \lambda\rvert=\lVert B\rVert$ (here $\lVert \cdot\rVert$ denotes the induced matrix norm from the Euclidean norm on $\mathbb C^n$). For definiteness suppose that $\lambda>0$ is an eigenvalue of multiplicity one, $Bx=\lambda x$, $\lVert B\lVert=\lambda$ and that all remaining eigenvalues of $B$ are contained in, say, $[-(1-\delta)\lVert B\rVert,(1-\delta)\lVert B\rVert]$ for some fixed positive $\delta$.

It is obvious that $\lVert A\rVert\le \lVert B\rVert$. I now first want to consider the extreme case when $\lVert A\rVert=\lVert B\lVert$. It follows that $A$ has an normalized eigenvector $y$ of eigenvalue $\pm\lambda$, $Ay=\pm\lambda y$ and we can compute \begin{align*} \lambda=\lvert\pm\lambda\rvert =\lvert\langle y, Ay\rangle\rvert= \Big\lvert\sum_{ab}\overline{y_a} A_{ab}y_b\Big\rvert\le \sum_{ab} \lvert y_a\rvert B_{ab} \lvert y_b\rvert =\langle\lvert y\rvert,B \lvert y\rvert\rangle\le \lambda =\langle x,Bx\rangle.\end{align*} In particular, it follows that $\lvert y\rvert =x$.

Now I am interested in how this can be made quantitative. For example, if $\lVert A\rVert \ge (1-\epsilon)\lVert B\rVert$, is it true that $\lVert \lvert y\rvert -x\rVert\le C \epsilon$ for some universal constant $C$ depending on the spectral gap of $B$? I think the above argument can be made quantitative to give $$\lVert \lvert y\rvert -x\rVert\le 2\sqrt{\frac{\epsilon}{\delta}},$$ where $\delta$ is the spectral gap of $B$ in the sense that $\big\lVert B\rvert_{x^\perp}\big\rVert\le (1-\delta)\lVert B\rVert$. I have the strong suspicion (supported by various numerical experiments) that the square-root bound in terms of $\epsilon$ is not optimal but that there is rather a linear bound.

I could imagine that $$ \lVert A\rVert=\lVert B\rVert\quad\text{if and only if}\quad A_{ab}=e^{i(\phi_a-\phi_b)} B_{ab}$$ for some $\phi\in\mathbb R^n$. (The if part is obvious because $x e^{i\phi}$ is an eigenvector of $A$ with eigenvalue $\lambda$)

Answer 1

It supprised me but your conjecture is true.

First we can suppose that the matrix $B$ is irrecductible. Otherwise we can just keep the $B$-stable subspace which contain the larger eigenvalue.

For $\|A\|\rightarrow \|B\|$, as in your remark we have $\||y|-x\|\rightarrow 0$. With $y_i=e^{i\phi_i}|y_i|$ And we replace $A_{ij}$ by $A_{ij}e^{i(\phi_i-\phi_j)}$ and $y_i$ by $|y_i|$. Therefore :$$ \|A\|=\sum_{ij}|y_i||y_j|A_{ij}|\rightarrow |\sum_{ij}x_ix_jB_{ij}|$$ and $|y_i|\rightarrow x_i>0$ implise that $\forall ij $ $A_{ij}$ have the same phase assymptotically. And therefore $A_{ij}\rightarrow B_{ij}$. So we proved your last statement and it is true for a small neibourhood.

We can then write $A_{ij}=e^{i\psi_{ij}}B_{ij}$ with $\psi_{ij} \rightarrow 0$. Because of unicity of the eigenvalue, $\|A\|=\lambda(\psi)$ and $y(\psi)$ are analytic in $\psi$ and we make the development. $$y(\psi)=x+r.\psi+h(\psi,\psi) +o(\|\psi^2\|)$$ $$\lambda(\psi)=\lambda+l.\psi+h_2(\psi,\psi)+o(\|\psi^2\|) $$ Because $\lambda$ is a maximum, $l=0$. Let $(f_i,\lambda_i)$ the eigenvectors, eigenvalues of $B$. ($\lambda_1=\|B\|,f_1 =x$). $A_{ij}=B_{ij}+i\psi_{ij}B_{ij}$ we have (https://fr.wikipedia.org/wiki/Théorie_de_la_perturbation_(mécanique_quantique)) $$ r.\psi = \sum_{k\neq 1} \frac{(f_k,(i\psi B)f_1)}{\lambda_1-\lambda_k} f_k$$ where $(\psi B)_{ij}=\psi_{ij} B_{ij}$. Because $f_k$ are real vector. Therefore $r.\psi$ is purely imaginary and there is no first linear term for $|y(\psi)|$! $$ \sum_{ij}|y_i||y_j|A_{ij}|=\sum_{ij}(x_i+\delta |y_i|)(x+\delta|y_j|)B_{ij}e^{i\psi_{ij}}|=\sum_{ij}(x_i+\delta |y_i|)(x+\delta|y_j|)B_{ij}(1-\psi^2_{ij})|+o(\|\psi\|^2)$$ with $\delta |y_i|=O(\|\psi\|^2)$.Because $\|x+\delta |y|\|=1$, $x\perp \delta |y|$ with ($o(\delta|y|$) and we have then $(\delta |y| B x)=O(\|\psi\|^3)$. And we can conclude $$ \sum_{ij}|y_i||y_j|A_{ij}|=\lambda-\sum_{ij}x_i x_j B_{ij}\psi_{ij}^2+o(\|\psi\|^2)$$ And we have proved that $\delta |y|=O(\|\psi \|^2)$ and $\lambda-\|A\|> c\|\psi\|^2$

Answer 2

This fails for pretty much the same reasons as in my comment above. Take something like $$ A = \begin{pmatrix} C & 0 \\ 0 & D \end{pmatrix} ,\quad D = \begin{pmatrix} 1 & 1 \\ 1 & -1 \end{pmatrix} $$ (everything in $A$ is a $2\times 2$ block), where $C$ has positive entries and $\|C\|=2-\epsilon$. This is then also the operator norm of $A$, since $\|D\|=\sqrt{2}$.

After taking absolute values, the eigenvalues of $D$ are $0,2$, so the operator norm has only increased by $\epsilon$. But the $y$'s with $\|Ay\|=\|A\| \|y\|$ (strictly speaking, you didn't even define $y$ in this more general situation, but this seems the only reasonable interpretation) are in $L(e_1,e_2)$, while $x\in L(e_3,e_4)$, so $\| |y|-x\|^2 = 2$.

A positive statement would have to take into account how far $\lambda$ is from the rest of $\sigma(B)$ (as you already hinted at yourself), and I think it's straightforward to derive such a statement, using your argument.