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.

  • $\begingroup$ The general pattern is that you have $-1$ under the diagonal and $1$ over the diagonal? $\endgroup$ – J. M. is a poor mathematician Aug 22 '11 at 3:19
  • $\begingroup$ In general it looks like $$\det(A_n)=\begin{cases}1\text{ if }n\text{ even}\\ 0\text{ if }n\text{ odd}\end{cases},$$ perhaps using cofactor expansion and splitting the problem into even and odd cases will help? $\endgroup$ – Zev Chonoles Aug 22 '11 at 3:25
  • $\begingroup$ I'll throw out my idea and hope somebody with better skills than me can finish it: $\begin{pmatrix}\textbf{E}&\textbf{F}\\ \textbf{G}&\textbf{H}\end{pmatrix}=\det\left(\textbf{H}-\textbf{G}\textbf{E}^{-1}\textbf{F}\right)\;\det\;\textbf{E}$; you can take $\mathbf H=\mathbf A_2$, $\mathbf E=\mathbf A_{2n-2}$ and $-\mathbf G^T=\mathbf F$ to be the $(2n-2)\times 2$ matrix whose entries are all $1$s. There seems to be a nice formula for the inverses; your problem reduces to proving that $-\textbf{F}^T\textbf{E}^{-1}\textbf{F}$ is the zero matrix... $\endgroup$ – J. M. is a poor mathematician Aug 22 '11 at 3:52
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    $\begingroup$ And all this without actually having a definition of $A_{2n}$ for $n\ge3$... :-) $\endgroup$ – Did Aug 22 '11 at 15:25
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    $\begingroup$ @J.M.: For your first question, yes. :-) Indeed, I have to "define" it. $\endgroup$ – Jack Aug 23 '11 at 3:42

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 $1/\lambda$.

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    $\begingroup$ +1 Nice. Effectively you first subtract the next to last row from the last, then you subtract the third row (from the bottom) from the on below that et cetera to get the matrix $PA_{2n}$. In the end you could also kill several pairs of 1s from the first row by adding some of the rows number 3, 5, 7,... from it. That way you only get a single non-zero term in the row expansion. $\endgroup$ – Jyrki Lahtonen Aug 22 '11 at 12:36
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    $\begingroup$ This is cool! It never occurred to me to multiply the original matrix with an appropriately constructed triangular matrix, but this is a wonderful method! Too bad I can only upvote once... $\endgroup$ – J. M. is a poor mathematician Aug 23 '11 at 5:28
  • $\begingroup$ @Jyrki Lahtonen: Yes, robjohn has added a new proof based on the same idea, but he use even indices to instead of odd indices. $\endgroup$ – user1551 Aug 24 '11 at 17:14

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 $$\sum_{\pi \in \Pi}\text{sgn}(\pi)=\text{sgn}(id)=1.$$

  • $\begingroup$ In order to prove that $\text{pf}(A)=1$ one can also use the expansion formula for pfaffians; see en.wikipedia.org/wiki/Pfaffian. $\endgroup$ – user26857 Nov 8 '12 at 15:48

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 $0$.

  • $\begingroup$ Looking back at user1551's post, my first step, getting to matrix $(1)$, is working with columns where user1551 worked with rows. My second step, getting to matrices $(2)$ and $(3)$, makes the computation of the determinant simpler. $\endgroup$ – robjohn Aug 24 '11 at 17:40

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 $\det(A_{2n}) = \det(A_2)$.


I will take the case of odd $n$ first, since my reasoning is similar but more complex in the even case.

For odd $n$, the fact that $\det A_n = 0$ follows from the fact that a skew-symmetric matrix of odd size is always singular, but here is another approach for the case in question:

Form the alternating sum of all the column vectors $v_1, v_2, \dots, v_n$ in the matrix $\sum_{i=1}^n (-1)^i v_i$.

The first row sum will be the sum $0 + 1 - 1 + 1 - \dots -1 = 0$ (an odd number of terms, one zero, which gives an even number of non-zero elements, half of them 1 and the other half -1).

An arbitrary row sum will likewise have an odd number of terms, one zero, thus an even number of non-zero terms, half of them 1 and the other half -1 (the zero is placed so that that the term before it is $(-1)^{k-1} (-1) = \pm 1$ and the element after it is $(-1)^{k+1} 1 = \mp 1$).

So $\sum_{i=1}^n (-1)^i v_i = 0$, the column vectors are linearly dependent. Therefore $\det A_n = 0$ for odd $n$.

Now for the even case.

Denote the column vectors in $A_n$ by $v_1, v_2, \dots, v_n$. For every distinct set of $n-1$ vectors $\{v_{i_1}, v_{i_2}, \dots, v_{i_{n-1}}\}$ where $i_1 < i_2 < \dots < i_{n-1}$ form an alternating sum: $$\sum_{k=1}^{n-1} (-1)^k v_{i_k}$$

Say that the vector $v_p$ is the one vector "missing" in this sum. Then, all row sums except for the $p$:th row will be 0, with the same reasoning as above; there will be an odd number of terms, exactly one will be 0, which implies an even number of non-zero terms, half of which will be 1 and the other half -1.

For the $p$:th row sum, there will be no zero terms, therefore the row sum will be either 1 or -1.

From this, it is obvious that $\det A = \pm 1$. However, as has been proven elsewhere on this site, the eigenvalues of an invertible real skew symmetric matrix comes in pairs of complex conjugates, $\lambda, \overline{\lambda}$. The determinant will therefore be $\lambda_1 \overline{\lambda_1} \cdots \lambda_{n/2} \overline{\lambda_{n/2}} = |\lambda_1|^2 \cdots |\lambda_{n/2}|^2$, which is a positive number.

We conclude that $\det A = 1$.

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    $\begingroup$ The determinant of any odd-dimensional skew-symmetric matrix is $0$. First, $\left|A^T\right|=\left|A\right|$. Since $n=\dim(A)$ is odd, and $A^T=-A$, we get that $$\left|A^T\right|=\left|-A\right|=(-1)^n\left|A\right|=-\left|A\right|$$ Thus, $\left|A\right|=-\left|A\right|$, so $\left|A\right|=0$. $\endgroup$ – robjohn Aug 22 '11 at 15:48

Let $A_{2n}$ the $2n\times 2n$ matrix defined by $(A_{2n})_{i,j} = \begin{cases} 1&\mbox{if }i<j\\ 0&\mbox{if }i=j\\ -1&\mbox{if } i>j \end{cases}$. We can show the result ($\det(A_{2n}) = 1$) by induction. The case $n=1$ and $n=2$ has been solved. Now, we can write $A_{2n+2}=\begin{pmatrix}A_{2n}&B\\-B^t &C\end{pmatrix}$ where $B$ is a $2n\times 2$ matrix whose entries are all $1$ and $C=\begin{pmatrix}0&1\\-1&0\end{pmatrix}$. Consider the line $l_{2n+2}$: we can transform it, computing $l_{2n+2}\leftarrow l_{2n+2}-l_{2n+1}$. It becomes $(0,\ldots,0,-1,-1)^t$ (the $2n$ first components are $0$). After that we make $c_{2n+2}\leftarrow c_{2n+2}-c_{2n+1}$: we get $(0,\ldots,0,1,0)$. Now we conclude, expanding by the last column and after the last row (the first expansion gives $(-1)^{2n-1+2n}=-1$ and the second $(-1)^{2n-1+2n-1}\cdot (-1)$), hence we get $\det A_{2n+2}=(-1)\cdot(-1)\cdot\det A_{2n} =\det A_{2n}$.


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