**Disclaimer.** This is a cross-post from [math.SE](https://math.stackexchange.com/questions/2416819/matrix-of-absolute-values-and-largest-eigenvector) 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 $\lambda$)



Entrywise modulus matrix and the largest eigenvector
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)](https://fr.wikipedia.org/wiki/Th%C3%A9orie_de_la_perturbation_(m%C3%A9canique_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 $\lambda-\|A\|> c\|\psi\|^2$


