Let $A$ be the adjacency matrix of an undirected and simple graph (so it is symmetric and the entries are 0 or 1).
Suppose I have a decomposition $A = B + C$ where $B$ and $C$ are matrices of same dimensions as $A$ and they also only contain 0 or 1.
My questions are
Can I say anything about the largest eigenvalue of a weighted sum of $B$ and $C$, say $bB + cC$ where $b$ and $c$ are constants?
Can I have any constraint for $b$ and $c$ so that the largest eigenvalue of $bB + cC$ less than 1?
Because $A = B + C$ is symmetric matrix, we can use several results to receive bounds on the eigenvalues of $A$ in terms of the eigenvalues of $B$ and $C$. Weyl's Inequalities state that for Hermitian matrices $B, C$ with eigenvalues ordering nondecreasingly $\lambda_1(B) \leq \lambda_2(B) \leq \cdots \leq \lambda_n(B)$ and $\lambda_1(C) \leq \lambda_2(C) \leq \cdots \leq \lambda_n(C)$, then $$ \lambda_i(B+C) \leq \lambda_{i+j}(B) + \lambda_{n-j}(C) \quad\text{for}\quad j = 0, 1, \ldots, n-i $$ and $$ \lambda_{i-j+1}(B) + \lambda_j(C) \leq \lambda_i(B+C) \quad\text{for}\quad j = 1, \ldots, i $$ for each $i = 1, \ldots, n$. Equality holds for a pair $i, j$ if and only if there is a nonzero vector $\mathbf x$ such that $A \mathbf x = \lambda_{i+j}(A) \mathbf x$ and $B \mathbf x = \lambda_j(B) \mathbf x$. If $A$ and $B$ have no common eigenvector, then these inequalities are strict.
Scaling a matrix by a factor of $c$ scales each eigenvalue by the same factor, so the same inequalities hold for linear combinations of $B$ and $C$. If you know more about your underlying graph it's easy to derive even stronger results. In particular, the adjacency matrices of degree-regular graphs always have the constant vector as an eigenvector, which gives them many nice properties.
Using Weyl's inequalities, it's possible to find bounds for $b$ and $c$ such that $\lambda_n(bB + cC) < 1$. If $b, c > 0$, this constraint is satisfied for $$\begin{align*} \lambda_n(bB) + \lambda_n(cC) &< 1 \\ b \lambda_n(B) + c \lambda_n(C) &< 1. \end{align*}$$ For $b, c < 0$ the sign of each eigenvalue is reversed and $\lambda_n(bB) \neq b \lambda_n(B)$, so the above does not hold.