Determinant of a special matrix The question is related to the eigenvalue problem. Using MAPLE, I calculated the following determinants:
$$\begin{align}
\begin{vmatrix}
   -\lambda & 1 \\
  1 & -\lambda \\
\end{vmatrix}&=\lambda^2-1\\
\begin{vmatrix}
  -\lambda & 1 & 0 \\
   1& -\lambda & 1 \\
   0& 1 & -\lambda \\
\end{vmatrix}&=-\lambda^3+2\lambda\\
\begin{vmatrix}
  -\lambda & 1 &0  &1  \\
   1& -\lambda & 1 &0  \\
   0&  1& -\lambda & 1 \\
   1&  0&  1& -\lambda \\
\end{vmatrix}&=\lambda^4-4\lambda^2\\
\begin{vmatrix}
  -\lambda &1  &0  &1  &0  \\
   1& -\lambda &1  &0  &1  \\
   0&  1& -\lambda &1  &0  \\
   1&  0&  1& -\lambda &1  \\
   0&  1&  0&  1& -\lambda \\
\end{vmatrix}&=-\lambda^5+6\lambda^3\\
\begin{vmatrix}
  -\lambda &1  &0  &1  &0  &1  \\
   1& -\lambda &1  &0  &1  &0  \\
   0&  1& -\lambda &1  &0  &1  \\
   1&  0&  1& -\lambda &1  &0  \\
   0&  1&  0&  1& -\lambda &1  \\
   1&  0&  1&  0&1 & -\lambda \\
\end{vmatrix}&=-9\lambda^4+\lambda^6\\
\end{align}
$$
But I have no idea how to calculate the determinants quickly by hand. Here is my question:  

What is the determinant in the $n$ by $n$ case?


Let us denote the matrix in $n$-dimensions by $\Lambda_n$ and the two non vanishing eigenvalues by $a$ and $b$. We have
$a+b = tr(\Lambda_n) = 0$
$a^2+b^2 = tr(\Lambda_n^2) = 2\lceil \frac{n}{2}\rceil\lfloor\frac{n}{2}\rfloor$
which implies that:
$a ^2= b^2 =-ab = \lceil \frac{n}{2}\rceil \lfloor\frac{n}{2}\rfloor$
Thus the two nonvanishing eigenvalues contributtion to the characteristic polynomial: 
$\lambda^2-\lceil \frac{n}{2}\rceil \lfloor\frac{n}{2}\rfloor$
