If the eigenvalues of $A_{m\times m}$ are $\lambda_1 \le \lambda_2 \le \lambda_3 \le \dotsb \le \lambda_m$ then find the eigenvalues of $B$ where $$B=\begin{bmatrix} C &D\\E&A\end{bmatrix}$$ where $$C=I_{k\times k}$$ and $$D=J_{k\times m},E=J^t$$ where $J$ is the all $1$ matrix.
Here $C,D,E,A $ are so arranged such that $B$ is a symmetric $(m+k) \times (m+k)$ matrix.
Is there any way I can find the eigenvalues of the above matrix?
Any help will be highly helpful.
Let $v$ be an eigenvector of $B$, and partition it as such: $$v=\begin{bmatrix} x\\y \end{bmatrix}$$ where I assume that $C$ is $n$ x $n$ and $x$ has $n$ rows; where as $A$ is $m$ x $m$ and $y$ has $m$ rows.
Then, $B v = \gamma v$, where $\gamma$ are the eigenvalues of $B$ to be determined. As mentioned earlier, there is a multiplicity of eigenvalues at $\gamma = 1$. To determine the rest, we assume that $\gamma \neq 1$. Doing some algebra gives:
$Cx + Jy = \gamma x$ and $J^tx + Ay = \gamma y$. Thus, $x = \frac{1}{\gamma-1}Jy$. (For completion, you can obtain the eigenvectors for $\gamma =1$ cases by finding the null space of $J^t$.) Substituting, we get: $$\frac{1}{\gamma-1}J^tJy + Ay -\gamma y = 0$$
Look at the eigenvector decomposition of $A = V\Lambda V^{-1}$, where $V=[v_1, v_2, ... v_n]$ and $\Lambda = diag(\lambda_1, \lambda_2, ... \lambda_n)$. Then, $$\frac{1}{\gamma-1}J^tJy + V\Lambda V^{-1}y -\gamma y = 0$$ Let $u=V^{-1}y$ and substituting: $$\frac{1}{\gamma-1}J^tJVu + V\Lambda u -\gamma Vu = 0$$ Multiplying on the left by $V^{-1}$, then $$\frac{1}{\gamma-1}V^{-1}J^tJVu + (\Lambda -\gamma I) u = 0$$
Notice that $J = \mathbf 1_n$ x $\mathbf 1_m^t$, and thus the above equation can be rewritten as $\frac{n}{\gamma-1}ab^tu + (\Lambda -\gamma I) u = 0$ where $a=V^{-1} \mathbf 1_m$ and $b^t=\mathbf 1_m^t V$ are column vectors. This implies $$\text{det} \left(\frac{n}{\gamma-1}ab^t + (\Lambda -\gamma I) \right)= 0$$ which can then be solved for $\gamma$.