A Property of the Determinant of a Circulant Matrix

 I've written about the circulant matrix only twice before in the following posts:

• The Circulant Matrix of a Number and its Permanent on the 1st January 2023
• Numbers and Matrices on the 15th December 2022

Today, with my diurnal age being 26975, I was struggling to find some interesting properties for this number and my thoughts fell to testing whether the sum of its digits divided the determinant of its circulant matrix. It did. I tested for other numbers and to my surprise this was always the case. What is going on?

Well, it all becomes clear when we consider generalised 2 x 2 circulant matrix (ignoring the trivial case of one digit numbers). Let's start with a generalised two digit number. Let's call it \(ab\). This number produces the circulant matrix shown in Figure 1.

Figure 1

This matrix has a determinant of \(a^2-b^2=(a+b)(a-b) \) and thus \(a+b\) will always divide it. Let's now consider a three digit number \(abc\) with sum of digits \(a+b+c\). It's circulant matrix is shown in Figure 2.

Figure 2

The determinant of this matrix is:$$a^3 - 3 a b c + b^3 + c^3\\ = (a + b + c) (a^2 - a b - a c + b^2 - b c + c^2)$$Once again, \(a+b+c\) will always divide this determinant. Let's look at a four digit number \(abcd\) with sum of digits \(a+b+c+d\). The circulant matrix is shown in Figure 3.

Figure 3

The determinant of this matrix is:$$a^4 - 4 a^2 b d - 2 a^2 c^2 + 4 a b^2 c + 4 a c d^2 - b^4 + 2 b^2 d^2 - 4 b c^2 d + c^4 - d^4\\=(a - b + c - d) (a + b + c + d) (a^2 - 2 a c + b^2 - 2 b d + c^2 + d^2)$$Once again, \(a+b+c+d\) will always divide this determinant. I won't go any further nor attempt a completely generalised proof but it seems apparent that the sum of the digits of any number will always divide the determinant of the number's circulant matrix. A surprising and interesting result.

Figure 4

I give credit to Wolfram Alpha for quickly calculating the determinants and factorising them.

Figure 5

In fact, I felt guilty after using Wolfram Alpha, because I knew SageMath could do the job just as well. So here is the latter's handling of the generalised five digit number \(abcde\). Figure 6 shows the number' circulant matrix:

Figure 6: permalink

Here is the determinant of this matrix:$$a^5 + b^5 - 5ab^3c + 5a^2bc^2 + c^5\\ + 5a^2b^2d - 5a^3cd - 5bc^3d + 5b^2cd^2 + 5ac^2d^2\\ - 5abd^3 + d^5 - 5a^3be + 5b^2c^2e - 5ac^3e\\ - 5b^3de - 5abcde + 5a^2d^2e - 5cd^3e + 5ab^2e^2\\ + 5a^2ce^2 + 5c^2de^2 + 5bd^2e^2 - 5bce^3 - 5ade^3 + e^5$$This factorises to:$$(a^4 - a^3b + a^2b^2 - ab^3\\ + b^4 - a^3c + 2a^2bc - 3ab^2c - b^3c\\ + a^2c^2 + 2abc^2 + b^2c^2 - ac^3 - bc^3\\ + c^4 - a^3d + 2a^2bd + 2ab^2d - b^3d\\ - 3a^2cd - abcd + 2b^2cd + 2ac^2d - 3bc^2d\\- c^3d + a^2d^2 - 3abd^2 + b^2d^2 + 2acd^2\\ + 2bcd^2 + c^2d^2 - ad^3 - bd^3 - cd^3 + d^4 \\- a^3e - 3a^2be + 2ab^2e - b^3e + 2a^2ce\\ - abce + 2b^2ce - 3ac^2e + 2bc^2e - c^3e\\ + 2a^2de - abde - 3b^2de - acde - bcde\\ + 2c^2de + 2ad^2e + 2bd^2e - 3cd^2e - d^3e\\ + a^2e^2 + 2abe^2 + b^2e^2 + 2ace^2 - 3bce^2\\ + c^2e^2 - 3ade^2 + 2bde^2 + 2cde^2 + d^2e^2\\ - ae^3 - be^3 - ce^3 - de^3 + e^4)\\(a + b + c + d + e)$$Clearly, once again, the sum of the digits \(a+b+c+d+e\) divides the determinant.

If we consider the permanent of the circulant matrix and not the determinant, then out of the first one million numbers there are 106343 numbers, roughly 10%, whose sum of digits divides the permanent (permalink).