Show that if $C^-$ is the generalized inverse of a symmetric matrix $C$ such that $C\mathbf1=0$ then $\dfrac{l'C^-l}{l'l}$ is constant for all vectors $l$ such that $l'\mathbf1=0$ iff $\theta C$ is idempotent for some $\theta>0$.
My attempts: I defined $x=\dfrac{l}{\sqrt{l'l}}$, then $x'C^-x$ is constant for all $x$ such that $x'\mathbf 1=0$ and $||x||=1$. But that doesn't seem to lead me to anywhere.
Let $\mathcal{C}(C)$ be the column space of $C$, respectively. First, $w'C\mathbf{1}=l'\mathbf{1}$ for some $w$. Thus, $Cw=l$. Since $l\in\mathcal{C}(C)$, $w=C^{-}l$. (I am not entirely sure about this step).
Let $\theta$ be a unique constant and $C=U\Lambda U'$ be the spectral decomposition of $C$ (which is guaranteed by symmetry). then: $$\theta=\frac{l'C^{-}l}{l'l}=\frac{w'CC^{-}Cw}{w'CCw}=\frac{w'Cw}{w'CCw}=\frac{y'\Lambda y}{y'\Lambda^2 y} $$
where $U'w=y$. For the equality to hold, $$\sum_{i}\lambda_{i}y_{i}^{2}=\sum_{i}\theta\lambda_{i}^{2}y_{i}^{2}$$ Thus, $eigen(C)=\{0,\theta^{-1}\}$, and $eigen(\theta C)=\{0,1\}$. Hence, $\theta C$ is idempotent for $\theta>0$.
Let $\theta >0$ such that $\theta C$ is idempotent. Note that $\theta$ is unique, otherwise the idempotence will be lost. Also, let $x$ be an arbitrary vector such that $Cx=l$. Then $$\theta^{2}CC=\theta C\\ \theta^{2}x'CCx=\theta x'Cx\\ \theta=\frac{x'Cx}{x'CCx}\\ \theta=\frac{l'C^{-}l}{l'l}$$ Finally, it is easy to show that $x'C\mathbf{1}=l'\mathbf{1}=0$, which concludes the proof.