$A=\left(a_{ij}\right)\in M_{n\times n}(\mathbb R)$ s.t. $a_{ij}=a_{ji}=\begin{cases}0,&i=j\\0\lor 1&i\ne j\end{cases}$
The eigenvalues of $A$ are $\lambda_1, \dots, \lambda_n$
I want to find an upper bound of $\sum |\lambda_j|$.
It will be best if someone can give an upper bound $\frac{1}{2}n^{\frac{3}{2}}$ roughly when n is large enough.(or even less than it)
Suppose there are $m$ nonzero entries in $A$. Then using Cauchy-Schwartz and the fact that here $m=\text{Tr}(A^2)=\sum_{i =1}^n \lambda_i^2$, you get the bound:
\begin{equation} \sum_{i =1}^n \vert\lambda_i\vert \leq \bigg(n\sum_{i =1}^n \lambda_i^2\bigg)^{1/2}=(nm)^{1/2}. \end{equation}
This doesn’t really use the fact the diagonal is zero. For the matrix with all off-diagonals equal to 1, this is off by a roughly $\sqrt{n}$ factor, so I’m not sure how good it is. If you allow the off-diagonal entries to be $-1$ as well, this bound is actually tight, as shown by Hao Huang using a signing of the adjacency matrix of the Boolean hypercube in the recent resolution of the Sensitivity Conjecture.