eigenvalues of certain block matrices This question inquired about the determinant of this matrix:
$$
\begin{bmatrix}
  -\lambda &1  &0  &1  &0  &1  \\
   1& -\lambda &1  &0  &1  &0  \\
   0&  1& -\lambda &1  &0  &1  \\
   1&  0&  1& -\lambda &1  &0  \\
   0&  1&  0&  1& -\lambda &1  \\
   1&  0&  1&  0&1 & -\lambda
\end{bmatrix}
$$
and of other matrices in a sequence to which it belongs.  In a comment I mentioned that if we permute the indices 1, 2, 3, 4, 5, 6 to put the odd ones first and then the even ones, thus 1, 3, 5, 2, 4, 6, then we get this:
$$
\begin{bmatrix}
-\lambda & 0 & 0 & 1 & 1 & 1 \\
0 & -\lambda & 0 & 1 & 1 & 1 \\
0 & 0 & -\lambda & 1 & 1 & 1 \\
1 & 1 & 1 & -\lambda & 0 & 0 \\
1 & 1 & 1 & 0 & -\lambda & 0 \\
1 & 1 & 1 & 0 & 0 & -\lambda
\end{bmatrix}
$$
So this is of the form
$$
\begin{bmatrix}
A & B \\ B & A
\end{bmatrix}
$$
where $A$ and $B$ are symmetric matrices whose characteristic polynomials and eigenvalues are easily found, even if we consider not this one case of $6\times 6$ matrices, but arbitrarily large matrices following the same pattern.
Are there simple formulas for determinants, characteristic polynomials, and eigenvalues for matrices of this latter kind?
I thought of the Haynesworth inertia additivity formula because I only vaguely remembered what it said.  But apparently it only counts positive, negative, and zero eigenvalues.
 A: 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 $A-B$.
A: 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.
A: 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.
A: Because the subblocks of the second matrix (let's call it $C$) commute i.e. AB=BA,  you can use a lot of small lemmas given, for example here. 
And also you might consider the following elimination: Let $n$ be the size of $A$ or $B$ and let,(say for $n=4$)
$$
T = \left(\begin{array}{cccccccc}
     1     &0     &0     &0     &0     &0     &0     &0\\
     0     &0     &0     &0     &1     &0     &0     &0\\
    -1     &1     &0     &0     &0     &0     &0     &0\\
    -1     &0     &1     &0     &0     &0     &0     &0\\
    -1     &0     &0     &1     &0     &0     &0     &0\\
     0     &0     &0     &0    &-1     &1     &0     &0\\
     0     &0     &0     &0    &-1     &0     &1     &0\\
     0     &0     &0     &0    &-1     &0     &0     &1
\end{array} \right)
$$
Then , $TCT^{-1}$ gives 
$$
\hat{C} = \begin{pmatrix}-\lambda &n &\mathbf{0} &\mathbf{1} \\n &-\lambda &\mathbf{1} &\mathbf{0}\\ & &-\lambda I &0\\&&0&-\lambda I \end{pmatrix}
$$
from which you can identify the upper triangular block matrix. The bold face numbers indicate the all ones and all zeros rows respectively.  $(1,1)$ block is the $2\times 2$ matrix and $(2,2)$ block is simply $-\lambda I$.
EDIT: So the eigenvalues are $(-\lambda-n),(-\lambda+n)$ and $-\lambda$ with multiplicity of $2(n-1)$. Thus the determinant is also easy to compute, via their product.
