Finding the eigenvalues of a $n\times n$ $A=(a_{ij})$ defined by $a_{ij}=1$ if $i+j=n+1$, $0$ elesewhere. If $A=(a_{ij})\in M_n(\mathbb{R})$ is a matrix defined by $a_{ij}=1$ if $i+j=n+1$, $0$ otherwise. Then what are the eigenvalues of $A$? 
\begin{bmatrix}
    0    & 0      & \ldots        & \ldots & \ldots        & 0      & 1     \\
    0      & 0    & \ldots        & \ldots & \ldots        & 1      & 0      \\
    \vdots & \vdots & \vdots &        &               & \vdots & \vdots \\
    \vdots & \vdots &               & \vdots &               & \vdots & \vdots \\
    \vdots & \vdots &               &        & \vdots & \vdots & \vdots \\
    0     & 1     & \ldots        & \ldots & \ldots        & 0    & 0      \\
    1     & 0      & \ldots        & \ldots & \ldots        & 0      & 0
\end{bmatrix}
I could find out that the determinant of $A$ is $(-1)^{n+1}$.

Note that, if $e_j=(0,0,\ldots,0,1,0,\ldots,0)$, where the $1$ is in the $j-$th place, then
$$
A(e_j+e_{n-j})=e_j+e_{n-j} \tag{1}
$$ 
while
$$
A(e_j-e_{n-j})=-(e_j-e_{n-j}) \tag{2}
$$
From $(1)$ we obtain $\lfloor\frac{n+1}{2}\rfloor$ linearly independent eigenvectors
corresponding to the eigenvalue $1$, while $(2)$ provides $\lfloor\frac{n}{2}\rfloor$ linearly independent eigenvectors corresponding to the eigenvalue $-1.$
Note that
$$
\left\lfloor\frac{n}{2}\right\rfloor+\left\lfloor\frac{n+1}{2}\right\rfloor=n,
$$
for all $n$.
So, the characteristic polynomial of $A$ is
$$
p_A(x)=-(-1)^n(x-1)^{\left\lfloor\frac{n}{2}\right\rfloor}(x+1)^{\left\lfloor\frac{n+1}{2}\right\rfloor}.
$$
