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Let $A_n$ be a matrix, odd dimension $n \times n$, constructed from the sequence of natural numbers in such a way that we begin the sequence from the upper left corner and next numbers are inserted into the matrix along antidiagonal direction, starting from the top of the matrix (examples below) .

For such matrix the determinant is calculated.

Surprisingly for the three examples below the value of determinant is equal to the central entry of the matrix with the modification of sign between two consecutive cases.

$\det \begin{bmatrix}1 & 2 & 4\\3 &\color{red} 5 & 7\\6 & 8 & 9\end{bmatrix} =-\color{red}5$

$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11\\3 & 5 & 8 & 12 & 16\\6 & 9 & \color{red}{13} & 17 & 20\\10 & 14 & 18 & 21 & 23\\15 & 19 & 22 & 24 & 25\end{bmatrix} =\color{red}{13} $

$\det \begin{bmatrix}1 & 2 & 4 & 7 & 11 & 16 & 22\\3 & 5 & 8 & 12 & 17 & 23 & 29\\6 & 9 & 13 & 18 & 24 & 30 & 35\\10 & 14 & 19 & \color{red}{25} & 31 & 36 & 40\\15 & 20 & 26 & 32 & 37 & 41 & 44\\21 & 27 & 33 & 38 & 42 & 45 & 47\\28 & 34 & 39 & 43 & 46 & 48 & 49\end{bmatrix} = -\color{red}{25} $

Question:

  • does the pattern continue for a greater $n$?

  • if so what is the explanation for the pattern?

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  • $\begingroup$ Note that the pattern holds for the often overlooked $1\times 1$ matrices as well: The determinant of $[\color{red}{1}]$ is $\color{red}{1}$. $\endgroup$
    – Arthur
    Apr 9, 2019 at 11:11
  • $\begingroup$ @Arthur Yes, 1 is also "central" in [1], and it has + sign, this change of sign is also strange.. $\endgroup$
    – Widawensen
    Apr 9, 2019 at 11:13
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    $\begingroup$ Here's how it continues: oeis.org/A069480. Your observation would make a nice comment, BTW. $\endgroup$ Apr 9, 2019 at 11:27
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    $\begingroup$ The formula section of the OEIS link does confirm this: It says that for odd $n$, the determinant is indeed an alternating multiplied by $\frac{(2n+1)^2}{2}$, which is the middle element of the corresponding matrix.. $\endgroup$
    – Arthur
    Apr 9, 2019 at 11:34
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    $\begingroup$ @Widawensen In this case, the easiest determinant to start with first might be the one which has central value $0$: we just have to show that this matrix is singular. But the fact that as you add $k$ to all entries, the determinant changes by $\pm k$, might also be hard to prove... $\endgroup$ Apr 9, 2019 at 22:28

1 Answer 1

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Let $M_n$ be the matrix whose determinant we want to know (the matrix where we fill in $n^2$ consecutive integers antidiagonally, starting at any value $k$), and let $x$ be its middle entry. For convenience, let $m = \frac{n-1}{2}$. (I will use $n=7$ in my example matrices, but my argument applies to all odd $n$.)

First, replace the rows $r_1, r_2, \dots, r_n$ of $M_n$ with the $n-1$ differences $r_2 - r_1$, $r_3 - r_2$, ..., $r_n - r_{n-1}$ followed by the middle row $r_{m+1}$. This is an invertible sequence of row operations, so we have: $$ \det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ x+a_1 & x+a_2 & x+a_3 & x & x-a_3 & x-a_2 & x-a_1 \end{bmatrix}. $$ where $x+a_1,\dots, x+a_m, x, x-a_m, \dots, x-a_1$ are the entries of the middle row of $M_n$ (which always has this symmetric form). We show that the determinant above is always equal to $(-1)^{m} x$.

By linearity of the determinant, we can write this as $$ \det(M_n) = \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ a_1 & a_2 & a_3 & 0 & -a_3 & -a_2 & -a_1 \end{bmatrix} + x \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix}. $$ The first determinant is $0$ by cofactor expansion along the last row. The cofactors of $a_i$ and $-a_i$ are equal for each $i$, since one submatrix can be obtained from the other by reversing the rows, then reversing the columns, so they cancel after multiplying one by $a_i$ and the other by $-a_i$.

For the second matrix, we first do a similar operation to columns. Keeping the first column $c_1$ and replacing every other $i^{\text{th}}$ column $c_i$ by the difference $c_i - c_{i-1}$, which is an invertible set of row operations, we have $$ \det\begin{bmatrix} 2 & 3 & 4 & 5 & 6 & 7 & 7 \\ 3 & 4 & 5 & 6 & 7 & 7 & 6 \\ 4 & 5 & 6 & 7 & 7 & 6 & 5 \\ 5 & 6 & 7 & 7 & 6 & 5 & 4 \\ 6 & 7 & 7 & 6 & 5 & 4 & 3 \\ 7 & 7 & 6 & 5 & 4 & 3 & 2 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{bmatrix} = \det\begin{bmatrix} 2 & 1 & 1 & 1 & 1 & 1 & 0 \\ 3 & 1 & 1 & 1 & 1 & 0 & -1 \\ 4 & 1 & 1 & 1 & 0 & -1 & -1 \\ 5 & 1 & 1 & 0 & -1 & -1 & -1 \\ 6 & 1 & 0 & -1 & -1 & -1 & -1 \\ 7 & 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 \end{bmatrix} $$ By expansion along the last row (which introduces a multiple of $(-1)^{n+1} = 1$) and after reversing the rows (which introduces a multiple of $(-1)^{m}$) we simplify this to $$(-1)^{m}\det\begin{bmatrix} 0 & -1 & -1 & -1 & -1 & -1 \\ 1 & 0 & -1 & -1 & -1 & -1 \\ 1 & 1 & 0 & -1 & -1 & -1 \\ 1 & 1 & 1 & 0 & -1 & -1 \\ 1 & 1 & 1 & 1 & 0 & -1 \\ 1 & 1 & 1 & 1 & 1 & 0 \end{bmatrix}. $$ Now things get a bit fancier. This $n-1 \times n-1 = 2m \times 2m$ matrix (call it $A$) is a skew-symmetric matrix, so its determinant is the square of its Pfaffian. We use the simplified formula given in the Wikipedia link above: the sum $$ \text{pf}(A) = \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha) a_{i_1j_1} a_{i_2j_2}\dotsm a_{i_mj_m} $$ where $\alpha$ runs over all pairings $\{(i_1,j_1), \dots, (i_m, j_m)\}$ with $i_k < j_k$ in each pair and $i_1 < \dots < i_m$; $\pi_{\alpha}$ is the permutation with $\pi_\alpha(2k-1) = i_k$ and $\pi_\alpha(2k) = j_k$. For our particular matrix, $a_{i_kj_k}$ is always $-1$, so $$ \text{pf}(A) = (-1)^m \sum_{\alpha \in \Pi} \text{sgn}(\pi_\alpha). $$ For each pairing $\alpha$, if we find the first index $k$ where $j_k > i_{k+1}$ and switch $j_k$ with $j_{k+1}$, we get another pairing $\beta$ with $\text{sgn}(\pi_\beta) = -\text{sgn}(\pi_\alpha)$; if we apply the same operation to $\beta$, we get $\alpha$ back. So this splits up the pairings into, uh, pairs of pairings whose terms in the Pfaffian sum cancel; the exception is the pairing $\{(1,2), (3,4), \dots, (2m-1,2m)\}$, for which no index $k$ exists. So $\text{pf}(A) = (-1)^m$ after all the cancellations, and therefore $\det(A) = 1$.

So we get $\det(M_n) = 0 + x \cdot (-1)^{m} = (-1)^{\frac{n-1}{2}}x$: plus or minus the middle entry of $M_n$.

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  • $\begingroup$ I'm very grateful for your partial answer - it has given me some new ideas to rethink... $\endgroup$
    – Widawensen
    Apr 10, 2019 at 8:21
  • $\begingroup$ I wonder whether in the case of the second determinat the same technique applied by you with substracting the rows could not lead to the further simplification.. $\endgroup$
    – Widawensen
    Apr 10, 2019 at 8:59
  • $\begingroup$ Anyway I have received now {{-1, -1, -1, -1, -1, 0, 1}, {-1, -1, -1, -1, 0 ,1 ,1}, {-1, -1 , -1 , 0 ,1, 1 ,1}, {-1, -1, 0 ,1 ,1 ,1, 1}, {-1, 0 , 1 ,1 ,1, 1, 1}, {5, 5, 4 ,3, 2 ,1, 0}, {1, 1, 1, 1, 1, 1, 1}} at which I'm stacked.. $\endgroup$
    – Widawensen
    Apr 10, 2019 at 9:25
  • $\begingroup$ I have also received for $5 \times 5$ such form $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\3 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ which has det =$-1 \ \ $. Interestingly $\begin{equation*}\left[\begin{matrix}0 & 0 & 0 & 1 & 2\\0 & 0 & 1 & 2 & 2\\0 & 1 & 2 & 2 & 2\\4 & 3 & 2 & 1 & 0\\1 & 1 & 1 & 1 & 1\end{matrix}\right]\end{equation*}$ has det = $0$. $\endgroup$
    – Widawensen
    Apr 10, 2019 at 11:53
  • $\begingroup$ Now I have a complete answer. $\endgroup$ Apr 10, 2019 at 22:11

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