Finding eigenvalues of an anti-diagonal matrix Let $A=(a_{ij}) \in M_{n}(\mathbb{R})$ such that
$$a_{ij}=
  \begin{cases} 
      \hfill 1    \hfill & \text{ if $i+j=n+1$} \\
      \hfill 0 \hfill & \text{ otherwise} \\
  \end{cases}      $$  
I am required to find determinant and eigenvalues for $n=10$ and $n=100$.
ATTEMPT
Eigenvalues for $n=10$ are just set containing elements $1$. So determinant is $1$. But the solution states it should be $-1$. How do I do for $n=100$ and the general case?
Thanks.
 A: Hint: your matrix is a permutation matrix. The characteristic polynomial of the permutation matrix corresponding to $\sigma\in\mathbf S_n$ is the product of the characteristic polynomials of (the matrices of) the disjoint cycles in the decomposition of $\sigma$, and since the companion matrix for the polynomial $X^d-1$ is the permutation matrix of a $d$-cycle, the characteristic polynomial of such a cycle is $X^d-1$. So all that is left is to find your permutation, its decomposition into disjoint cycles, and to combine the roots of the corresponding factors $X^d-1$ into a big multiset of eigenvalues.
A: here is way to compute the eigenvalues and eigenvectors. we will look at the cases $n = 2, 4, \cdots$ to see a pattern. i will use the standard basis $e_1, e_2, \cdots, e_n$ where $e_j$ has $1$ at the $j$th  entry and $0$ everywhere else.
let us look at the case $n= 2.$ we have $$Ae_1 = e_2,\, Ae_2 = e_1 \tag 1$$
adding and subtracting the two equations, we get $$A(e_1+e_2) = e_1 + e_2,\ A(e_1 - e_2)=-(e_1-e_2) \tag2 $$ hat is, the eignevalues of $A$ are $\pm 1$ and the corresponding eigenvectors are $e_1\pm e_2.$
now the case $n = 4.$  we have $$Ae_1 = e_4,\, Ae_2 = e_3,\  Ae_3 = e_2,\ Ae_4 = e_1\tag 2$$
 by pairing the first and the fourth, second and the third and using the case for $n = 2,$  we get the eigenvalues and the eigenvectors of $A$ are
$$\pm 1, \ \pm 1, \text{ corresponding eigenvectors are }e_1 \pm e_4 \text{ and } e_2 \pm e_3.$$
i hope you see how to extend this result to $n= 10$ and $n = 100.$
A: Hint product of eigenvalues is determinant of a particular matrix while sum of trace of a matrix is the sum of eigenvalues. See now we can create a principal diagonal by changing $C_{10} to C_{1}$  and similarly changing till $C_{5} to C_{6}$ so now you get a digonal matrix . Whose $det(A)=1$ but as we have changed $5$ times columns so its $(-1)^5.det(A)=-1.1=-1$ hope you can think for $n=100$
