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Consider an $n\times n$ matrix $M_n$ where the sequence $$\{1,2,3,\dots,n^2\} \mod 4=\{1,2,3,0,1,2,3,\dots\}$$ forms a clock-wise spiral, in that given order. For example, $$M_4=\begin{bmatrix} 1&2&3&0\\ 0&1&2&1\\ 3&0&3&2 \\ 2&1&0&3 \end{bmatrix} \qquad \text{and} \qquad M_5=\begin{bmatrix} 1&2&3&0&1\\ 0&1&2&3&2 \\ 3&0&1&0&3 \\ 2&3&2&1&0 \\ 1&0&3&2&1 \end{bmatrix}.$$

Question. Is it true that $$\det(M_{2n})=3(2n-1)4^{n-1} \qquad \text{and} \qquad \det(M_{2n+1})=-(3n^2-1)4^n\,\,\,?$$

Added clarification. To understand the construction of the above matrices, take a look at the matrices from my other MO question. Then, reduce the entries modulo $4$ and follow through by computing the determinants.

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  • $\begingroup$ Something I noticed when doing some larger versions: each column is the same as the column 4 before, with 2 exceptions, one of which is 2, the other of which is -2. Other than the first four columns, and switching the signs of some columns, you get 2s and -2s in two diagonals. $\endgroup$
    – user44191
    May 24, 2017 at 5:07
  • $\begingroup$ By quotient matrix, we can compute the determinants of the above examples simply. Maybe, by some good partitioning of your matrices, we can prove your claim. $\endgroup$
    – Shahrooz
    May 24, 2017 at 19:43
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    $\begingroup$ Seems like $M_n$ is the central submatrix of $M_{n+2}$. I don't know how to use this though. $\endgroup$
    – François Brunault
    May 26, 2017 at 10:58
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    $\begingroup$ If $C_1,\ldots,C_{2n}$ denote the columns of $M_{2n}$ then $C_i \equiv C_{i+2} \pmod{2}$. Replacing $C_{2i+1}$ by $C_{2i+1}-C_1$ and $C_{2i+2}$ by $C_{2i+2}-C_2$ for any $1 \leq i \leq n-1$ yields a matrix where $2n-2$ columns consist of even entries, hence the determinant is divisible by $4^{n-1}$. Probably a similar argument can show that the determinant is divisible by 3. $\endgroup$
    – François Brunault
    May 30, 2017 at 15:22
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    $\begingroup$ Adding the first and last rows (resp. columns) in $M_{2n}$ if $n \equiv 0$ resp. $1 \pmod{2}$ shows that $\det M_{2n}$ is divisible by 3. $\endgroup$
    – François Brunault
    May 30, 2017 at 21:12

1 Answer 1

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Yes, it is true. More generally, the entries $1,2,3,0$ can be replaced by arbitrary numbers $a,b,c,d$, in which case the determinant of $M_n$ can be computed in terms of the four numbers $u = d-b$, $v = a-c$, $U = d+b$ and $V = a+c$ as follows:

  • If $n=4k$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{4} v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left( 2k-1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) . $$

  • If $n=4k+2$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =-\dfrac{1}{4}% v^{n-4}\left( v^{4}-u^{2}v^{2}+\left( U^{2}-V^{2}\right) \left( \left( 2k+1\right) ^{2}v^{2}-\left( 2k\right) ^{2}u^{2}\right) \right) . $$

  • If $n=4k+1$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{2} u^{n-3}\left( u^{2}\left( v+V\right) -\left( 2k\right) ^{2}v\left( U^{2}-V^{2}\right) \right) . $$

  • If $n=4k+3$ for some positive integer $k$, then $$ \det\left( M_{n} \right) =\dfrac{1}{2} vu^{n-3}\left( u^{2}+vV-\left( 2k+1\right) ^{2}\left( U^{2}-V^{2}\right) \right) . $$

As you can imagine, this is not very fun to prove. I have a writeup (The 4-periodic spiral determinant) in which I attempt at making the idea clear without going into all the details; in particular, annoying computations are relegated to SageMath and to the reader (and on some occasions to a combination of both). Even at that level of terseness, it is 24 pages long. I would normally hope that something nicer can be found, but with the complexity of the answer I am not too hopeful.

The proof starts out as suggested by @user44191 in one of the comments to the original post; thus the matrix is brought to a form where all entries are zero except for those in northwesternmost $4\times 4$-submatrix and on four sub-antidiagonals (namely, the $1$-st, the $3$-rd, the $5$-th and the $7$-th sub-antidiagonals) below the main antidiagonal. Then, I turn the matrix upside down, so that the sub-antidiagonals become the super-diagonals. I then perform Laplace expansion with respect to the last $4$ rows. All $4 \times 4$-minors from the last $4$ rows can be explicitly computed (only $\dbinom{7}{4}$ of them nonzero, and this can be further reduced by looking at the vanishing of the complementary minors), so it remains to compute the complementary $\left(n-4\right)\times\left(n-4\right)$-minors. For this, Jacobi's complementary minor theorem turns out to be of use, along with an explicit computation of the inverse of a certain power series.

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