Determinant of an $n\times n$ Toeplitz matrix Let $A = (a_{ij}) \in R^{n\times n}$. Find the determinant if:
$$a_{ij}= |i-j|$$
So we have the symmetric matrix
\begin{bmatrix}
0 & 1 & 2 & 3 & 4 & \dots & n-1 \\
1 & 0 & 1 & 2 & 3 & \dots & n-2 \\
2 & 1 & 0 & 1 & 2 & \dots & n-3 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \dots & \dots & \dots & 0
\end{bmatrix}
But i can't find a way to diagonalize the matrix nor find the determinant by Laplace expansion.
Any ideas ??
 A: If we do row reduction, i.e. $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_2-R_1$, and then again do $R_n-R_{n-1}$, $R_{n-1}-R_{n-2}$, up to $R_3-R_2$ and finally $R_2+R_1$, then we get
\begin{align}
&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-2 \\
2 & 1 & 0 & 1 & 2 & \cdots & n-3 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \cdots & \cdots & \dots & 0
\end{vmatrix}\\ \ \\
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & -1 & -1 & -1 & -1 & \cdots & -1 \\
1 & 1 & -1 & -1 & -1 & \cdots & -1 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
1&1&1&\cdots&\cdots&-1&-1 \\
1 & 1 & 1 & \cdots & \cdots & 1 & -1
\end{vmatrix}\\ \ \\
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&2&0&0 \\
0 & 0 & 0 & \cdots & \cdots & 2 & 0
\end{vmatrix}\\
\end{align}
Now we calculate the determinant and its minors all by the last row:
\begin{align}
=&\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&2&0&0 \\
0 & 0 & 0 & \cdots & \cdots & 2 & 0
\end{vmatrix}\\ \ \\
=&-2\begin{vmatrix}
0 & 1 & 2 & 3 & 4 & \cdots & n-1 \\
1 & 0 & 1 & 2 & 3 & \cdots & n-1 \\
0 & 2 & 0 & 0 & 0 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0&0&0&\cdots&\cdots&2&0 \\
\end{vmatrix}\\ \ \\
&=(-1)^{n-2}2^{n-2}\begin{vmatrix}0&n-1\\1&n-1 \end{vmatrix}\\ \ \\
&=(-1)^{n-1}2^{n-2}(n-1).
\end{align}
A: Call your matrix $A$ and let $J$ be the upper triangular nilpotent Jordan block of size $n$ (i.e. the superdiagonal of $J$ contains ones and all other entries are zero). Then $A=U+U^T$, where $U=J+2J^2+\ldots+(n-1)J^{n-1}$. Now, note that $(I-J)U=J+J^2+\ldots+J^{n-1}$ is the strictly upper triangular matrix of ones. Therefore
$$
(I-J)U(I-J^T)=
\pmatrix{
-1&&&&1\\
  &-1&&&1\\
    &&\ddots&&1\\
      &&&-1&1\\
         &&&&0
}
$$
and hence
\begin{align*}
B &:= (I-J)A(I-J^T) = (I-J)(U+U^T)(I-J^T) \\
&= (I-J)U(I-J^T) + ((I-J)U(I-J^T))^T \\
&=
\pmatrix{
-2&&&&1\\
  &-2&&&1\\
    &&\ddots&&1\\
      &&&-2&1\\
         1&1&\cdots&1&0
}.
\end{align*}
Since $\det(I-J)=1$, we get $\det B=\det A$. Now, adding one half of each of the first $n-1$ rows of $B$ to the last row, we get
$$
\det B=
\det\pmatrix{
-2&&&&1\\
&-2&&&1\\
&&\ddots&&1\\
&&&-2&1\\
&&&&\frac{n-1}2
}=\frac{(-2)^{n-1}(n-1)}2.
$$
