Let $A = [a\_{ij}]\_{n\times n}$ be a Hermitian matrix, such that $|a\_{ij}| =1$ for $i \neq j$, and $a\_{ii} = 0$ for each $i$.
I am interested in a tight lower bound of $\|A\|\_\*:=\sum\_{i=1}^n |\lambda\_i(A)|$, where $\lambda\_i(A)$'s are eigenvalues of $A$.


Note that, 
by minimizing $\sum\_{i=1}^n |\lambda\_i|$ over two constraints $\sum\_{i=1}^n \lambda\_i = 0$ and $\sum\_{i=1}^n \lambda\_i^2= n(n-1)$, one can obtain $\sqrt{2n(n-1)}$ as a lower bound. But it seems that isn't tight.


On the other hand, if $A := J - I$ (all ones matrix minus identity), then $\sum\_i |\lambda\_i(A)| = 2(n-1)$. 


Is it true that $2(n-1)$ is actually a lower bound (for large enough matrices, say $n \geq 10$) ?


**Remarks:**


* As Alex's answer below, the minimum of trace norm of such matrices may be less than $2(n-1)$, even for arbitrarily large matrices.
* But, as a comment of @fedja, the minimum is $(2+o(1))n$ as $n\to\infty$.


**Added:**


* In the particular case, when $a\_{ij}=±1$, the lower bound holds. See [this answer](https://mathoverflow.net/a/309214/53059) below, for an overview of the proof.



Is this lower bound for a norm of some complex matrices true?
No, it is not; in fact, $2(n-1)$ is a local *maximum*.


Let $B$ be a Hermitian matrix such that $|B\_{ij}|=1$ and $B\_{ii}=1$. We denote its eigenvalues by $\mu$ (not to confuse them with eigenvalues of $A$). It is easy to see that always $\mu\le n$: if $(x\_1,\dots,x\_n)$ is an eigenvector and $|x\_i|=\max\_j|x\_j|$ then
$$|\mu x\_i|=\left|\sum\_j B\_{ij}x\_j\right|\le \sum\_j |x\_j|\le n|x\_i|.$$


If $A=B-I$ then
$$\sum\_i |\lambda\_i|=\sum\_i |\mu\_i-1|.$$
In the case $B=J$ we have $\mu\_1=n$ and $\mu\_i=0,2\le i\le n$, hence
$$\sum\_i |\lambda\_i|=2(n-1).$$
If $B$ is not much different from $J$ then we still have one large eigenvalue $\mu\_1$ and plenty of small ones, $\mu\_i<1,2\le i\le n$. In this more general case
$$\sum\_i |\lambda\_i|=\mu\_1-1+\sum\_{i=2}^n(1-\mu\_i)=2(\mu\_1-1)\le 2(n-1).$$ 
Naturally, even in a vicinity of $J$ it won't always be an equality.


