Determinant of a special skew-symmetric matrix

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.

Solution 1:

Here is a combinatorial way to answer this. If we have a skew-symmetric matrix $A=\{a_{ij}\}_{1\le i,j\le 2n}$, then $\det(A)=Pf(A)^2$, where $Pf(A)$ is the Pfaffian of $A$. We know from standard methods that $$Pf(A)=\sum_{\pi \in \Pi}\text{sgn}(\pi)a_{\pi(1),\pi(2)}\cdots a_{\pi(2n-1),\pi(2n)}$$ where $\Pi$ is the set of permutations $\pi\in S_{2n}$ which satisfy $\pi(2k-1)<\pi(2k)$ for $1\le k\le n$ and $\pi(1)\le \pi(3)\le \cdots \le \pi(2n-1)$. In our case all $a_{ij}$ with $i < j$ have the same value $-1$, so we only need to prove that $$|\sum_{\pi \in \Pi}\text{sgn}(\pi)|=1.$$ To do this we will exhibit an involution on $\Pi\backslash\{id\}$ (the permutations in $\Pi$ that are not the identity).

Let $\pi \in \Pi\backslash\{id\}$, there will be a smallest $k$ so that $\pi(2k-1)= \pi(2k+1)-1$. define $\pi'$ to be the same as $\pi$ but with $\pi'(2k)=\pi(2k+2)$ and $\pi'(2k+2)=\pi(2k)$. I will leave it as an exercise for you to prove that $\pi'\in \Pi\backslash\{id\}$, $\pi''=\pi$ and that $\text{sgn}(\pi')=-\text{sgn}(\pi)$ so that $$\sum_{\pi \in \Pi}\text{sgn}(\pi)=\text{sgn}(id)=1.$$

Solution 2:

Let $$P=\begin{pmatrix}1\\-1&1\\&-1&1\\&&\ddots&\ddots\\&&&-1&1\end{pmatrix}.$$ Then $$PA_{2n}=\begin{pmatrix}0&1&1&\ldots&1\\-1&-1\\&-1&-1\\&&\ddots&\ddots\\&&&-1&-1\end{pmatrix}.$$ Computing by row expansion, we get $\det(PA_{2n}) = 0 - (-1) + (-1) - \ldots + (-1) - (-1) = 1$. Since $\det(P)=1$, we are now done.

Edit: By considering $PA_nP^{-1}$, actually we can further show that the characteristic polynomial of $A_n$ is $p(\lambda)=\det(\lambda I_n-A_n)=\frac12\left((\lambda+1)^n+(\lambda-1)^n\right)$, regardless of whether $n$ is even or odd. Therefore, if $n$ is even and $\lambda$ is a (necessarily nonzero) eigenvalue of $A_n$, so is $1/\lambda$.

Solution 3:

Based upon J.M.'s comments, I'd like to approach this in a different way from Davide's answer. Starting from a slightly different partitioning $$A_{2n+2} = \left(\begin{array}{cc} A_2 & B \\ -B^T & A_{2n}\end{array}\right)$$ where $B$ is a $2\times 2n$ matrix with all entries set to $1$, we know

$$\det(A_{2n+2}) = \det(A_2)\det(A_{2n} + B^T A_2^{-1} B).$$

Inspection reveals that $A_2^{-1} = A_2^T$, so

$$A_2^{-1} B = \left( \begin{array}{rrc} -1 & -1 & \ldots \\ 1 & 1 & \ldots \end{array} \right)$$

which means $B^T A_2^{-1} B = 0$. Hence, $\det(A_{2n+2}) = \det(A_{2n})\det(A_2)$. Since we know the determinant of $A_{2n}$ for $n=1\ \text{and}\ 2$ is $1$, clearly $\det(A_{2n}) = 1 \ \forall\ n \ge 1$.

Edit: it occurs to me that the inductive step is simplified by recognizing that $\det(A_{2n+2}) = \det(A_{2n})$ because $\det(A_2) = 1$. Then, this implies $\det(A_{2n}) = \det(A_2)$.

Solution 4:

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 $0$.