Let $A \in \mathbb{R}^{n \times n}$ be a symmetric matrix whose rows and columns sum to one. $A$ is not necessarily a doubly stochastic matrix, because negative entries are possible.
What can be said about the largest eigenvalue $\lambda$ of $A$? Is there a "good" upper bound for $\lambda$?
Additional constraint: Suppose that $|a_{ij}| \leq 1$. Does $\lambda \leq 1$ hold?
When the entries of $A$ are unbounded, $\lambda_\max(A)$ is unbounded too and hence there isn't any constant upper bound (but of course, you may use any submultiplicative matrix norm of $A$ as a non-constant upper bound). E.g. when $A=I+k\pmatrix{1&-1\\ -1&1}$ with $k\ge0$, we have $\lambda_\max(A)=2k+1$.
When the entries of $A$ have moduli $\le1$, we have $\lambda_\max(A)\le\rho(A)\le\|A\|_\infty\le n$. When $n$ is odd, this upper bound is attainable by the checkerboard matrix $A$ with $a_{ij}=(-1)^{i+j}$. When $n$ is even, if we set $A=C\oplus1$ where $C$ is the checkerboard matrix of size $n-1$, then $\lambda_\max(A)=n-1$. Therefore, there always exists a symmetric generalised stochastic matrix $A$ with $\lambda_\max(A)\ge n-1>1$ for every $n\ge3$.
