If $\theta>0$ and $C_{n\times n}$ is a symmetric matrix of rank $n-1$ (with $\vec{\mathbf 1}$ as the only vector in kernel) such that $$\theta C^2=C$$ How to show that all the diagonal entries of $C$ are equal and so offdiagonal entries of $C$ are also equal?
My attempts: I tried to show that $C=aI+b\mathbf 1\mathbf1'$ for some numbers $a$ and $b$ but failed. Also tried to show that $e_i'Ce_i=e_j'Ce_j$ for all $i$ and $j$, but couldn't prove why $||Ce_i||=||Ce_j||$ (where $e_i$ is the canonical basis vector). Please help.
Hints. Since $C$ is symmetric and $\mathbf1$ is an eigenvector of $C$ corresponding to the zero eigenvalue, $C$ has an orthonormal eigenbasis $\{u_1,u_2,\ldots,u_n\}$ such that $u_1=\mathbf1/\sqrt{n}$ and $C=\sum_{i\ge2}\lambda_iu_iu_i^T$. Apply the condition $\theta C^2=C$ to relate $\theta$ to $\lambda_i$ for each $i\color{red}{\ge2}$. Hence use the equality $I=\sum_{i\color{red}{=1}}^nu_iu_i^T$ to show that $C=aI+b\mathbf1\mathbf1^T$ for some $a$ and $b$.