Sum of divisors and unitary divisors as the eigenvalue and the spectral norm of some addition matrix? 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? 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.

*

*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 $U_n$.


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


*Is $V_n$ a subgroup of $U_n$?


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