Let $n$ be a natural number and $D\_n$ be the set of divisors.
We can make this set to a ring by observing that each divisor $d$ has


$$0 \le v\_p(d) \le v\_p(n)$$


Hence we can add two divisors $d,e$ by setting:


$$d \oplus e := \prod\_{p | n} p^{v\_p(d)+v\_p(e) \mod (v\_p(n)+1)}$$


and similarily we can multiply them by setting:
$$d \otimes e := \prod\_{p | n} p^{v\_p(d) \cdot v\_p(e) \mod (v\_p(n)+1)}$$


Then, if $n = p\_1^{a\_1} \cdots p\_r^{a\_r}$, this ring will be isomorphic to the ring


$$\mathbb{Z}/(a\_1+1) \times \cdots \times \mathbb{Z}/(a\_r+1)$$


If $n$ is squarefree, than this reduces to :


$$d\oplus e = \frac{de}{\gcd(d,e)^2}$$


$$d\otimes e = \gcd(d,e)$$


and the ring of divisors is a boolean ring as defined here
[Boolean ring of unitary divisors / Structure of unitary divisors?](https://mathoverflow.net/questions/369751/boolean-ring-of-unitary-divisors-structure-of-unitary-divisors) and here
<https://math.stackexchange.com/questions/3799607/does-this-characteristic-polynomial-factor-into-linear-factors-over-the-integers/3799759>


If we consider the addition table ($\oplus$) of this ring as a matrix, than it is clear that the sum of divisors $\sigma(n)$ is an eigenvalue to the eigenvector:


$$(1,\cdots,1)$$


Here is as an example the addition ($\oplus$) table for $n=12$:


$$\left(\begin{array}{rrrrrr}
1 & 2 & 3 & 4 & 6 & 12 \\
2 & 4 & 6 & 1 & 12 & 3 \\
3 & 6 & 1 & 12 & 2 & 4 \\
4 & 1 & 12 & 2 & 3 & 6 \\
6 & 12 & 2 & 3 & 4 & 1 \\
12 & 3 & 4 & 6 & 1 & 2
\end{array}\right)
$$


I have checked numerically ($n=1,\cdots,60$) that


$$\sigma(n) = |A\_n^k|\_2^{1/k}, \forall k \ge 1$$


where $A\_n$ is the addition matrix of this ring.


1. *Is there a proof for this last equality (where $|.|\_2$ denotes the spectral norm)?* (This question is proved here: <https://math.stackexchange.com/questions/3800389/ring-of-divisors-of-a-natural-number-and-the-sum-of-divisors-as-an-eigenvalue-an> )


Similarliy we can make the set $U\_n$ of unitary divisors into a boolean ring by setting:


$$a\oplus b = \frac{ab}{\gcd(a,b)^2}$$


$$a \otimes b = \gcd(a,b)$$


I have checked numerically similarliy to the above ($\sigma^\*(n) = $ sum of unitary divisors):


$$\sigma^\*(n) = |B\_n^k|\_2^{1/k}, \forall k \ge 1$$


where $B\_n$ is the addition matrix of $U\_n$.


2. Can this be proven?


To each eigenvalue $\lambda$ with eigenvector $v\_{\lambda}$ of $B\_n$ 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:


$$\lambda = \sum\_{v \in V\_{\lambda}} v - \sum\_{u \in V\_{\lambda}^C} u$$


3. Is $V\_n$ a subgroup of $U\_n$?
4. Is $\lambda$ equal to the right hand side of the last equality?


Thanks for your help!


Related questions:


<https://math.stackexchange.com/questions/3800389/ring-of-divisors-of-a-natural-number-and-the-sum-of-divisors-as-an-eigenvalue-an>


[Boolean ring of unitary divisors / Structure of unitary divisors?](https://mathoverflow.net/questions/369751/boolean-ring-of-unitary-divisors-structure-of-unitary-divisors)


<https://math.stackexchange.com/questions/3799607/does-this-characteristic-polynomial-factor-into-linear-factors-over-the-integers/3799759>



Sum of divisors and unitary divisors as the eigenvalue and the spectral norm of some addition matrix?
In both cases you are really only using the additive structure of your rings, so this is really a question about abelian *groups*.


Assuming $n = p\_1^{a\_1} \cdots p\_r^{a\_r}$, when studying $A\_n$ we are working with the abelian group $$G=\mathbb{Z}/(a\_1+1)\mathbb Z \times \cdots \times \mathbb{Z}/(a\_r+1)\mathbb Z.$$
We can think of the elements of $G$ as tuples $s=(s\_1,\dots,s\_r)$ where $s\_i\in \mathbb Z/(a\_i+1)\mathbb Z$. Here $A\_n$ coincides with the group matrix $(x\_{s+t})\_{s,t\in G}$ where $x\_{(s\_1,s\_2,\dots,s\_r)}$ is set equal to the unique positive divisor of $n$ that satisfies $\nu\_{p\_i}(x)=s\_i\pmod{a\_i+1}$ for all $1\le i\le r$.


Now, $A\_n$ is a symmetric matrix so we are really just trying to prove that the spectral radius is $\sigma(n)=\sum\_{s\in G}x\_s$. The good news is that we can say way more: we can write down all eigenvalues of this matrix.


Let's define vectors $\mathbf v(\chi)$ indexed by irreducible characters of $G$, to be given by $\mathbf v(\chi)\_{s}=\chi(s)$. Then we can check that the following holds
$$A\_n\mathbf v(\chi)=\left(\sum\_{s\in G}\chi(s)x\_s\right)\mathbf v(\bar{\chi}).\tag{\*}$$
If we let $\lambda\_{\chi}=\sum\_{s\in G}\chi(s)x\_s$, then the eigenvalues of $A\_n$ are either equal to $\lambda\_{\chi}$ for some $\chi$ that is equal to it's own conjugate, or equal to $\pm \sqrt{\lambda\_{\chi}\lambda\_{\bar{\chi}}}$ for some $\chi$ that is not equal to it's own conjugate.


To prove this fact notice that $(\*)$ tells us that when $\chi =\bar{\chi}$ we have $\mathbf v(\chi)$ as an eigenvector with eigenvalue $\lambda\_{\chi}$, and when $\chi\neq \bar{\chi}$ we see that $A\_n$ acts as
$$\begin{pmatrix}0 & \lambda\_{\chi}\\
\lambda\_{\bar{\chi}} & 0\end{pmatrix}$$
on the span of $\{\mathbf v(\chi),\mathbf v(\bar{\chi})\}$.


Finally it remains to notice that since all $\chi(s)$ are roots of unity, the largest eigenvalue is $\sum\_{s\in G}x\_s$ corresponding to the trivial character.


For $B\_n$ you can repeat a similar argument but for the group
$$G=(\mathbb Z/2\mathbb Z)^r$$
with group matrix $(x\_{s+t})\_{s,t\in G}$ and $x\_{(s\_1,\dots,s\_r)}$ chosen to be the unique positive divisor of $n$ that satisfies
$$\nu\_{p\_i}(x)=\begin{cases}
0 & \text{if } s\_i=0\\
a\_i & \text{otherwise},
\end{cases}.$$
This also gives positive answers to your last two questions because every character of $G$ in this case is $\pm 1$ valued, and the elements where the value is $+1$ form a subgroup.


If you want to read more about such type of results you can look up K. Konrad's notes [The origin of representation theory](https://kconrad.math.uconn.edu/articles/groupdet.pdf) which focus on the related group matrix $(x\_{s-t})\_{s,t\in G}$, and how understanding its determinant/spectrum for (first abelian and later general) groups began the study of representation theory.


