Let $A\in \lbrace 0,1\rbrace^{n,n}$ be a symmetric matrix with $diag(A)=0$.
Suppose there exists $i$ and $j$ such that
$$\; \forall k\not\in \{i,j\}: \; A_{ik}\geq A_{jk} \quad (*)$$
$Let \; f: \lbrace 0,1\rbrace^{n,n} \to \mathbb{R}^{n,n}$ and $B=f(A)$
What i am looking for are functions $f$ such that one of the following two statements holds:
$(i) \;\quad (*) \implies B_{ik}\geq B_{jk} \quad \forall k=1,\dots,n$
$(ii) \quad (*) \implies \sum\limits_k B_{ik} \geq \sum\limits_k B_{jk}$
What are requirements for $f$ such that one of the two holds?
My current approach is using the spectral decomposition $A=X\Lambda X^T$ where $X=[x_1\vert x_2 \vert \dots \vert x_n]$ are the eigenvectors of $A$ and $\Lambda=diag(\lambda_1,\lambda_2,\dots,\lambda_n)$ are the corresponding eigenvalues. Then we can write $(*)$ as $$ \sum\limits_{l=1}^n x_i(l)\lambda_l x_k(l) \geq \sum\limits_{l=1}^n x_j(l)\lambda_l x_k(l)$$ and $$B=f(A)=Xf(\Lambda)X^T$$ and therefore $$B_{ik}= \sum\limits_{l=1}^n x_i(l)f(\lambda_l) x_k(l)$$
I think with these representations of $A$ and $B$, it should be possible to derive properties of $f$ such that either (i) or (ii) holds, but i can't figure it out