I am studying equiangular line. I a using the reference: https://www.sciencedirect.com/science/article/pii/0021869373901233
$G=\begin{bmatrix}
1 & \dots & \pm a\\
& \ddots & \vdots\\
\pm a & & 1
\end{bmatrix}$
$A=\frac {1}{\alpha}\left(G-I\right)=\begin{bmatrix}
0 & \dots & \pm 1\\
& \ddots & \vdots\\
\pm 1 & & 0
\end{bmatrix}$
- At page 495, why A has the smallest eigenvalue 0 or multiplicity v-r? I didn't know why the smallest eigenvalue is 0 and why its multiplicity is v-r. Additionally, degree of characteristic polynomial p( $\lambda$ ) is r,right?(here r is the dimension of vector space $R^{r}$) I didn't understand "G has order v, is symmetric, and has the smallest eigenvalue 0 of multiplicity
v - r. Therefore, A has the smallest eigenvalue -l/$\alpha$ of multiplicity v - r. These multiplicities equal v-r if the unit vectors span R, . " this paragraph. Could someone explain more about it?
2.The proof of Theorem3.4 says that A has the smallest eigenvalue -(1/$\alpha$) with multiplicity m where m$\ge$m-r.
I don't know why the smallest eigenvalue is -(1/ $\alpha$ ), and I don't know why multiplicity m where m$\ge$m-r ?