Does this characteristic polynomial factor into linear factors over the integers?

Question

Let $n$ be a natural number, $U_n := \{ d | d \text{ divides } n, \gcd(d,n/d)=1\}$ be the set of unitary divisors.

We can make $U_n$ to a boolean ring:

$$a \oplus b := \frac{ab}{\gcd(a,b)^2} = \frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}$$ and $$a \otimes b := \gcd(a,b)$$

Why does it seem that the characteristic polynomial of the addition ($\oplus$) table factors over the integers in linear factors?

1 x - 1
2 (x - 3) * (x + 1)
3 (x - 4) * (x + 2)
4 (x - 5) * (x + 3)
5 (x - 6) * (x + 4)
6 (x - 12) * (x - 2) * (x + 4) * (x + 6)
7 (x - 8) * (x + 6)
8 (x - 9) * (x + 7)
9 (x - 10) * (x + 8)
10 (x - 18) * (x - 4) * (x + 6) * (x + 12)
11 (x - 12) * (x + 10)
12 (x - 20) * (x - 6) * (x + 10) * (x + 12)
13 (x - 14) * (x + 12)
14 (x - 24) * (x - 6) * (x + 8) * (x + 18)
15 (x - 24) * (x - 8) * (x + 12) * (x + 16)
16 (x - 17) * (x + 15)
17 (x - 18) * (x + 16)
18 (x - 30) * (x - 8) * (x + 10) * (x + 24)
19 (x - 20) * (x + 18)
20 (x - 30) * (x - 12) * (x + 18) * (x + 20)
21 (x - 32) * (x - 12) * (x + 16) * (x + 24)
22 (x - 36) * (x - 10) * (x + 12) * (x + 30)
23 (x - 24) * (x + 22)
24 (x - 36) * (x - 14) * (x + 18) * (x + 28)
25 (x - 26) * (x + 24)
26 (x - 42) * (x - 12) * (x + 14) * (x + 36)
27 (x - 28) * (x + 26)
28 (x - 40) * (x - 18) * (x + 24) * (x + 30)
29 (x - 30) * (x + 28)
30 (x - 72) * (x - 24) * (x - 16) * (x - 12) * (x + 8) * (x + 24) * (x + 36) * (x + 48)

Here is the addition table for $n=2,6,30$:

$$ \left(\begin{array}{rr} 1 & 2 \\ 2 & 1 \end{array}\right) \left(\begin{array}{rrrr} 1 & 2 & 3 & 6 \\ 2 & 1 & 6 & 3 \\ 3 & 6 & 1 & 2 \\ 6 & 3 & 2 & 1 \end{array}\right) \left(\begin{array}{rrrrrrrr} 1 & 2 & 3 & 5 & 6 & 10 & 15 & 30 \\ 2 & 1 & 6 & 10 & 3 & 5 & 30 & 15 \\ 3 & 6 & 1 & 15 & 2 & 30 & 5 & 10 \\ 5 & 10 & 15 & 1 & 30 & 2 & 3 & 6 \\ 6 & 3 & 2 & 30 & 1 & 15 & 10 & 5 \\ 10 & 5 & 30 & 2 & 15 & 1 & 6 & 3 \\ 15 & 30 & 5 & 3 & 10 & 6 & 1 & 2 \\ 30 & 15 & 10 & 6 & 5 & 3 & 2 & 1 \end{array}\right) $$

Edit: If possible I would like to understand where the eigenvalues come from. Here is what I have so far:

Sketch of the idea:

To each eigenvalue $\lambda$ with eigenvector $v_{\lambda}$ we can associate a stabilizer group $V_{\lambda} \le U_n$:

$$V_{\lambda} = \{u \in U_n| \left < (u\oplus u_1,\cdots,u \oplus u_r)^T ,v_{\lambda}\right >=\lambda \}$$

Then it seems that:

$$\lambda = \sum_{v \in V_{\lambda}} v - \sum_{u \in V_{\lambda}^C} u$$

which would prove the whole thing and explain how the eigenvalues occur.

Here are some examples: $$n$$ $$\lambda, V_{\lambda}, \lambda$$

1
1 [1] 1
2
3 [1, 2] 3
-1 [1] -1
3
4 [1, 3] 4
-2 [1] -2
4
5 [1, 4] 5
-3 [1] -3
5
6 [1, 5] 6
-4 [1] -4
6
12 [1, 2, 3, 6] 12
2 [1, 6] 2
-4 [1, 3] -4
-6 [1, 2] -6
7
8 [1, 7] 8
-6 [1] -6
8
9 [1, 8] 9
-7 [1] -7
9
10 [1, 9] 10
-8 [1] -8
10
18 [1, 2, 5, 10] 18
4 [1, 10] 4
-6 [1, 5] -6
-12 [1, 2] -12
11
12 [1, 11] 12
-10 [1] -10
12
20 [1, 3, 4, 12] 20
6 [1, 12] 6
-10 [1, 4] -10
-12 [1, 3] -12
13
14 [1, 13] 14
-12 [1] -12
14
24 [1, 2, 7, 14] 24
6 [1, 14] 6
-8 [1, 7] -8
-18 [1, 2] -18
15
24 [1, 3, 5, 15] 24
8 [1, 15] 8
-12 [1, 5] -12
-16 [1, 3] -16
16
17 [1, 16] 17
-15 [1] -15
17
18 [1, 17] 18
-16 [1] -16
18
30 [1, 2, 9, 18] 30
8 [1, 18] 8
-10 [1, 9] -10
-24 [1, 2] -24
19
20 [1, 19] 20
-18 [1] -18
20
30 [1, 4, 5, 20] 30
12 [1, 20] 12
-18 [1, 5] -18
-20 [1, 4] -20

Answer 1

Here's a sketch of a proof by induction on the number $k$ of prime factors of $n$:

For $k=0$ we have $n=1$ and the fact is clear. If $k=1$ then $n$ is a prime power and $|U_n|=2$, and we have $$p_n(X)=\det\begin{pmatrix}1-X&n\\ n&1-X\end{pmatrix}=(1-X)^2-n^2=(1+n-X)(1-n-X),$$ which shows that the characteristic polynomial $p_n(X)$ of the addition table of $U_n$ splits into linear factors over the integers.

Now suppose that $k>1$. Let $q\in U_n$ be the largest prime power unitary divisor and let $m=\tfrac{n}{q}$. Then by induction hypothesis the characteristic polynomial $p_m(X)$ of the addition table of $U_m$ splits into linear factors over the integers.

We can rearrange the rows and columns of the addition table of $U_n$ so that the top left quarter of the table is precisely the addition table of $U_m$. Note that this only changes the characteristic polynomial by a factor of $\pm1$, so this does not affect the factorization. The rearranged addition table of $U_n$ is a block matrix of the form $$\begin{pmatrix}A&B\\C&D\end{pmatrix},$$ where $A=D$ is the addition table of $U_m$ and $B=C=qA$. It follows that \begin{eqnarray*} p_n(X)&=&\det\begin{pmatrix}A-XI&B\\C&D-XI\end{pmatrix} &=&\det\begin{pmatrix}A-XI&qA\\qA&A-XI\end{pmatrix}. \end{eqnarray*} Because $A-XI$ and $qA$ commute it follows that the determinant of this $2\times2$-block matrix equals \begin{eqnarray*} p_n(X)&=&\det((A-XI)^2-(qA)^2)\\ &=&\det\big((1+q)A-XI)((1-q)A-XI)\big)\\ &=&\det\big((1+q)A-XI\big)\cdot\det\big((1-q)A-XI\big)\\ &=&(1+q)^{2^{k-1}}\det(A-\tfrac{X}{1+q}I)\cdot(1-q)^{2^{k-1}}\det(A-\tfrac{X}{1-q}I)\\ &=&(1+q)^{2^{k-1}}p_m(\tfrac{X}{1+q})\cdot(1-q)^{2^{k-1}}p_m(\tfrac{X}{1-q}). \end{eqnarray*} By induction hypothesis $p_m(X)$ splits into linear factors over the integers, and hence so do $$(1+q)^{2^{k-1}}p_m(\tfrac{X}{1+q})\qquad\text{ and }\qquad (1-q)^{2^{k-1}}p_m(\tfrac{X}{1-q}).$$ It follows that also $p_n(X)$ splits into linear factors over the integers.