TITLE: Determinant of a special skew-symmetric matrix
QUESTION [10 upvotes]: Simple calculation show that:
$$
\begin{align}
\det(A_2)=\begin{vmatrix} 
   0& 1  \\
   -1& 0  
\end{vmatrix}&=1\\
\det(A_4)=\begin{vmatrix} 
   0& 1 &1 &1 \\
   -1& 0 &1&1\\
   -1& -1& 0&1\\
   -1& -1& -1&0 
\end{vmatrix}&=1
\end{align}
$$
Here is my question:

Is it true that $\det(A_{2n})=1$ for all $n\in{\mathbb Z_+}$?

With MAPLE, I tried some large $n$. And I guess it is true. But temporarily I have no idea how to show it.

REPLY [4 votes]: Adding a multiple of any column to another does not change the determinant of the matrix.  Therefore, for each $k$ from $n$ to $2$, subtract column $k-1$ from column $k$. This leaves us with a matrix that looks like this:
$$
\left[\begin{array}{r}
0&-1&0&0&\dots&0&0&0&0\\
1&-1&-1&0&\dots&0&0&0&0\\
1&0&-1&-1&\dots&0&0&0&0\\
1&0&0&-1&\dots&0&0&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
1&0&0&0&\dots&-1&-1&0&0\\
1&0&0&0&\dots&0&-1&-1&0\\
1&0&0&0&\dots&0&0&-1&-1\\
1&0&0&0&\dots&0&0&0&-1\\
\end{array}\right]\tag{1}
$$
Except in column $1$, matrix $(1)$ has $-1$s on the diagonal and the superdiagonal.
Since $n$ is even, we can add the even columns to column $1$ in matrix $(1)$ and get
$$
\left[\begin{array}{r}
-1&-1&0&0&\dots&0&0&0&0\\
0&-1&-1&0&\dots&0&0&0&0\\
0&0&-1&-1&\dots&0&0&0&0\\
0&0&0&-1&\dots&0&0&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&0&\dots&-1&-1&0&0\\
0&0&0&0&\dots&0&-1&-1&0\\
0&0&0&0&\dots&0&0&-1&-1\\
0&0&0&0&\dots&0&0&0&-1\\
\end{array}\right]\tag{2}
$$
Matrix $(2)$ has $-1$s on the diagonal and the superdiagonal. Matrix $(2)$ has determinant $(-1)^n=1$ since $n$ is even.
If $n$ were odd, then we could add the other odd columns to column $1$ in matrix $(1)$ and get
$$
\left[\begin{array}{r}
0&-1&0&0&\dots&0&0&0&0\\
0&-1&-1&0&\dots&0&0&0&0\\
0&0&-1&-1&\dots&0&0&0&0\\
0&0&0&-1&\dots&0&0&0&0\\
\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\\
0&0&0&0&\dots&-1&-1&0&0\\
0&0&0&0&\dots&0&-1&-1&0\\
0&0&0&0&\dots&0&0&-1&-1\\
0&0&0&0&\dots&0&0&0&-1\\
\end{array}\right]\tag{3}
$$
Column $1$ of matrix $(3)$ is all $0$s; therefore, its determinant is $0$.