$A$ and $B$ are two square symmetric matrices, each of order $n \times n.$ Furthermore $\text{Rank}(A)=\text{Rank}(B)=n-1$. Also $A\textit 1=\textit 0$ and $B\textit 1=\textit 0$.
Let $C=A+B$.
I want to know about $\text{Rank}(C)$, specifically if $\text{Rank}(C)$ can be $n-1$. That is, under which condition $\text{Rank}(C)=\text{Rank}(A)=\text{Rank}(B)$.
I know that $\text{Rank}(C)\leq \text{Rank}(A)+\text{Rank}(B),$ but here this implies $\text{Rank}(C) \leq 2(n-1)$. This is of no help for $n\geq 2$ since $C$ being a $n×n$ matrix, $\text{Rank}(C)\leq n$ anyway.
Any suggestion is appreciated. Thanks in advance.