How to find the eigenvalues of the following block matrix Suppose $n \times n$ matrix $A$ is symmetric, where $n \in \mathbb N$. If the eigenvalues of $A$ are denoted by $\lambda_1\le \lambda_2\le \dots \le \lambda_n$, find the eigenvalues of the following $(2n+1) \times (2n+1)$ matrix in terms of $A$
$$M=\begin{bmatrix}
A_{n\times n} & & A_{n\times n} && \mathbf{0}_{n\times 1}
\\
A_{n\times n} && \mathbf{0}_{n\times n} && J_{n\times 1}\\
\mathbf{0}_{1\times n}&& J_{1\times n} && 0
\end{bmatrix}$$
where $J$ is the matrix with all entries equal to $1$.
What I did was I tried taking an eigenvector of $M$ with the corresponding eigenvalue as $\rho$ and did a straight forward calculation.
But the problem is I am not getting any way to express the eigenvalues of $M$ in terms of eigenvalues of $A$.
Is there any particular trick to solve these type of problems?
Can someone please help me out?

You cannot express the eigenvalues of $M$ in terms of the eigenvalues of $A$.
Consider the special case where $A$ is invertible. If the eigenvalues of $M$ depend only on the eigenvalues of $A$, then so does $\det(M)=\det(A)^2(e^TA^{-1}e)$, where $e$ denotes the vector of ones. In turn, $e^TA^{-1}e$ depends only on the eigenvalues of $A$, but this is clearly false unless $A$ is a scalar multiple of the identity matrix.
Generally speaking, it is rarely the case that one can express the eigenvalues of a matrix in terms of the eigenvalues of the matrix's sub-blocks.
