eigenvalues of certain block matrices

We have $$ \det \left( \begin{array}{cc} A & B\\ C & D \end{array} \right) = \det(A-BD^{-1}C) \det(D), $$ where the matrix $A-BD^{-1}C$ is called a Schur complement. In your case, $A=D=-\lambda I_n$ and $B=C=J_n$ = the order $n$ matrix with all entries equal to 1. So, the RHS is equal to $\det(-\lambda I_n + \frac{n}{\lambda} J_n) \det(-\lambda I_n) = (-n)^n \det(-\frac{\lambda^2}{n}I_n + J_n)$. If I remember correctly, $\det(xI_n + J_n) \equiv x^{n-1}(x+n)$, but you should check whether this is true or not.

I am not sure whether I understand what you want to ask.. but the following are some facts on the matrix of this type

$\det\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}=\det(A+B)\det(A-B)$. The eigenvalues of $\begin{bmatrix} A & B \\\\ B & A \end{bmatrix}$ are the union of eigenvalues of $A+B$ and the eigenvalues of $A-B$.

Your $2n\times 2n$ matrix $M$ acts on the vector space $V=\mathbb C^n\oplus\mathbb C^n$. Now if $W_1=\{(v,v):v\in\mathbb C^n\}$ and $W_2=\{(v,-v):v\in\mathbb C^n\}$, then we also have $V=W_1\oplus W_2$. Moreover, both $W_1$ and $W_2$ are invariant under $M$, so to find the eigenvalues/eigenvectors/characteristic polynomial/etc, it is enough to do it for those restrictions: they are $A+B$ and $A-B$.

This way you obtain, for example, the facts mentioned in Sunni's answer immediately.