Symmetric and anti-symmetric matrices and maximal eigenvalues

Question

Suppose we start with a symmetric $n \times n$ matrix $A$, the elements of which are either $1$ or $0$. All the diagonal elements of this matrix are set to be $0$. So, $\lambda_{\text{max}}=\sup \limits_{||v||=1} \langle v^{T}Av \rangle , v\in \mathbb R^n$.

Now, randomly change $1$'s to $-1$ so that the new matrix $A_1$ is anti-symmetric. Hence, $H=iA_1$ is Hermitian, the eigenvalues of which will be real and symmetrically distributed about zero. Call the maximum eigenvalue of this $\eta_{\text{max}}$. By randomly generating such matrices I have found that we have $\lambda_{\text{max}} \geq \eta_{\text{max}}$.

Using perturbation we can see that, $\lambda_{\text{max}} \xrightarrow{A \rightarrow H} \eta=\lambda_{\text{max}}-\langle v^{T} \Delta v\rangle +\mathcal O(\Delta^2) $. Here, $\Delta=A-H$ is also Hermitian with elements in $\{1 \pm i,0\}$ and all components of $v$ are non-negative ($v$ is an eigenvector corresponding to $\lambda_{\text{max}}$). So, the perturbation is $2\sum_{i>j} v_iv_j\sigma_{i,j}$ where $\sigma$ are either $1$ or $0$. Hence, this is positive. But this argument as the higher order terms may change things.

Naively, it seems obvious because some of the elements of $A$ are now becoming negative. But, is there a way to rigorously prove this?

Here, I add a random simulation backing this claim. Reds and Greens are $\lambda_{\text{max}}$ and $\eta_{\text{max}}$ respectively.

Answer 1

If $iA_1v=\eta v$, $\|v\|=1$, then $w=(|v_1|,\ldots ,|v_n|)$ is still normalized and $\langle w, Aw\rangle \ge \langle v, iA_1 v\rangle =\eta$, so the claim follows from min-max.