Vtyjkyuk

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

Here $n=pq,p<q$ and $p,q $ are primes and $l=phi(n)+1$.

$$Q=beginbmatrix(n-1)I_ltimes l&&&&&& J_ltimes n-l \J^T_(n-l)times l&&&&&& A_(n-l)times (n-l) endbmatrix$$

$A$ is a diagonal matrix of the form

$$A=beginbmatrix
C_(q-1)times (q-1) & 0 \
0 & D_(p-1)times (p-1)
endbmatrix$$

where $$C= beginbmatrix
p(q-1) & 1 & 1 &ldots & 1\1 & p(q-1) & 1 & ldots & 1\1 & 1 & p(q-1) &ldots & 1 \ ldots &ldots& ldots & ldots & ldots \ ldots & ldots & ldots& ldots & ldots \ ldots& ldots& ldots & ldots &ldots
\1 &1 &1 &ldots & p(q-1)
endbmatrix$$

and $$D=
beginbmatrix
q(p-1) & 1 & 1 &ldots & 1\1 & q(p-1) & 1 & ldots & 1\1 & 1 & q(p-1) &ldots & 1 \ ldots &ldots& ldots & ldots & ldots \ ldots & ldots & ldots& ldots & ldots \ ldots& ldots& ldots & ldots &ldots
\1 &1 &1 &ldots & q(p-1)
endbmatrix$$

where $J$ is the all $1$ matrix.

Here $C=J_stimes s+big(p(q-1)-1big)I_stimes s$ and $D=J_ttimes t+big(q(p-1)-1big)I_ttimes t$, where $s+t=n-l$. Therefore, the eigenvalues of $C$ and $D$ are
$p(q-1)-1$ (multiplicity $colorreds-1$), $n+p(q-1)-1$ (multiplicity $1$) and $q(p-1)-1$ (multiplicity $colorredt-1$), $n+q(p-1)-1$ (multiplicity $1$), respectively which are also the eigenvalues of $A$.

Now, consider $Q$ in the following form
$$Q=beginbmatrix(n-1)I & J_ltimes s & J_ltimes t\ J_stimes n-l & C & mathbf0\J_ttimes n-l & mathbf0 & Dendbmatrix.$$
Let $Cx_i=lambda_i x_i$ and $Dy_j=mu_jy_j$, where $lambda_i=p(q-1)-1$, $x_iperp mathbf1_s$ and $mu_j=q(p-1)-1$, $y_jperp mathbf1_t$.
Observe that $beginbmatrixmathbf0\x_i\mathbf0endbmatrix$ and $beginbmatrixmathbf0\mathbf0\y_jendbmatrix$ are eigenvectors of $Q$ corresponding to the eigenvalues $lambda_i$ and $mu_j$, respectively. The other eigenvalues should be of the form $X=beginbmatrixmathbf1_l\k_1mathbf1_s\k_2mathbf1_tendbmatrix$, for some constants $k_1$ and $k_2$.
From $QX=lambda X$, we get the following system of equations
$$begincasesn-1+k_1s+k_2t=lambda,\n-l+k_1(n+pq-p-1)=k_1lambda,\n-l+k_2(n+pq-q-1)=k_2lambda.endcases$$
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 $Q$.