Find the eigen values of $Q$ 
Find all eigen values of the following matrix $Q$:

Here $n=pq,p<q$ and $p,q $ are primes and $l=\phi(n)+1$.
$$Q=\begin{bmatrix}{(n-1)I}_{l\times l}&&&&&& J_{l\times n-l} \\J^T_{(n-l)\times l}&&&&&&   A_{(n-l)\times (n-l) }\end{bmatrix}$$
$A$ is a diagonal matrix of the form 
$$A=\begin{bmatrix}
C_{(q-1)\times (q-1)} & 0 \\
0 & D_{(p-1)\times (p-1)}
\end{bmatrix}$$
where $$C=  \begin{bmatrix}
p(q-1) & 1 & 1 &\ldots & 1\\1 & p(q-1) & 1 & \ldots  & 1\\1 & 1 & p(q-1) &\ldots & 1 \\ \ldots &\ldots& \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots& \ldots & \ldots \\ \ldots& \ldots& \ldots & \ldots &\ldots
\\1 &1 &1 &\ldots & p(q-1)
\end{bmatrix}$$
and $$D=
\begin{bmatrix}
q(p-1) & 1 & 1 &\ldots & 1\\1 & q(p-1) & 1 & \ldots  & 1\\1 & 1 & q(p-1) &\ldots & 1 \\ \ldots &\ldots& \ldots & \ldots & \ldots \\ \ldots & \ldots & \ldots& \ldots & \ldots \\ \ldots& \ldots& \ldots & \ldots &\ldots
\\1 &1 &1 &\ldots & q(p-1)
\end{bmatrix}$$
where $J$ is the all $1$ matrix.
 A: Here $C=J_{s\times s}+\big(p(q-1)-1\big)I_{s\times s}$ and $D=J_{t\times t}+\big(q(p-1)-1\big)I_{t\times t}$, where $s+t=n-l$. Therefore, the eigenvalues of $C$ and $D$ are 
$p(q-1)-1$ (multiplicity $\color{red}{s-1}$), $n+p(q-1)-1$ (multiplicity $1$) and $q(p-1)-1$ (multiplicity $\color{red}{t-1}$), $n+q(p-1)-1$ (multiplicity $1$), respectively which are also the eigenvalues of $A$. 
Now, consider $Q$ in the following form
$$Q=\begin{bmatrix}(n-1)I & J_{l\times s} & J_{l\times t}\\ J_{s\times n-l} & C  & \mathbf{0}\\J_{t\times n-l} & \mathbf{0} & D\end{bmatrix}.$$ 
Let $Cx_i=\lambda_i x_i$ and $Dy_j=\mu_jy_j$, where $\lambda_i=p(q-1)-1$, $x_i\perp \mathbf{1}_s$ and $\mu_j=q(p-1)-1$, $y_j\perp \mathbf{1}_t$. 
Observe that $\begin{bmatrix}\mathbf{0}\\x_i\\\mathbf{0}\end{bmatrix}$ and $\begin{bmatrix}\mathbf{0}\\\mathbf{0}\\y_j\end{bmatrix}$ are eigenvectors of $Q$ corresponding to the eigenvalues $\lambda_i$ and $\mu_j$, respectively. The other eigenvalues should be of the form $X=\begin{bmatrix}\mathbf{1}_l\\k_1\mathbf{1}_s\\k_2\mathbf{1}_t\end{bmatrix}$, for some constants $k_1$ and $k_2$.
From $QX=\lambda X$, we get the following system of equations
$$\begin{cases}n-1+k_1s+k_2t=\lambda,\\n-l+k_1(n+pq-p-1)=k_1\lambda,\\n-l+k_2(n+pq-q-1)=k_2\lambda.\end{cases}$$ 
Eliminating $k_1$ and $k_2$ from the above system of equation, you will get a cubic equation in $\lambda$. Then, substituting the relationship between $n, p, q, s, t$, you will get a better simplified equation whose roots are the rest three eigenvalues of $Q$.
