What are the eigenvalues of $(a_m+a_n)^2$? Let $a_m\in \mathbb{R}$ and consider the matrix defined by $A_{mn} = (a_m+a_n)^2$. Using python and inputting a random vector $a = (a_n:a_n\in \mathbb{R})$ of various length and various values, the eigenvalues of $A$ always seem to consist of 1 negative value and 2 positive values. All the other eigenvalues are $0$.
Based on this, it seems that there should be a formula which determines the 3 nontrivial eigenvalues of $A$ (and possibly all the eigenvectors of $A$), but I can't think of a rigorous proof.
EDIT: If $a_n$ had some "symmetry" so that $\sum a_n =\sum a_n^3=0$, then it's easy to see that $a_n$  is an eigenvector and that $a_n^2 \pm c$ are eigenvectors where
$$
c^2 = \frac{1}{N}\sum a_n^4
$$
However, the statement seems to be more general. Also, I'm not sure why all the other eigenvalues are 0.
 A: First, the matrix definitely has rank at most $3$, since it is the sum of three matrices
$$
A = \begin{pmatrix}
a_1^2 & a_1^2 & \cdots & a_1^2 \\
a_2^2 & a_2^2 & \cdots & a_2^2 \\
\vdots & \vdots & \ddots & \vdots \\
a_n^2 & a_n^2 & \cdots & a_n^2
\end{pmatrix}
+2
\begin{pmatrix}
a_1a_1 & a_1a_2 & \cdots & a_1a_n \\
a_2a_1 & a_2a_2 & \cdots & a_2a_n \\
\vdots & \vdots & \ddots & \vdots \\
a_na_1 & a_na_2 & \cdots & a_na_n
\end{pmatrix}
+
\begin{pmatrix}
a_1^2 & a_2^2 & \cdots & a_n^2 \\
a_1^2 & a_2^2 & \cdots & a_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
a_1^2 & a_2^2 & \cdots & a_n^2
\end{pmatrix}
$$
each of which has rank at most $1$.
Let's assume some genericity going forward, which is necessary to avoid more degenerate outcomes (like the zero matrix, for instance). I'll assume that $a_1\ne0$ and that $a_1,a_2,a_3$ are distinct.
Since it is a symmetric matrix, $A$ is positive (or negative) semidefinite if and only if all of its leading principal minors are nonnegative (or nonpositive). Note that the first leading principal minor is simply the positive quantity $a_1^2$, which rules out the possibility that $A$ is negative semidefinite. On the other hand, (my computer tells me that) the third leading principal minor is the negative quantity $-2(a_1-a_2)^2(a_1-a_3)^2(a_2-a_3)^2$, which rules out the possibility that $A$ is positive semidefinite.
That gets us down to two possibilities: one negative and two positive eigenvalues (what the OP observes), or one positive and two negative eigenvalues. I don't know how to finish from here. I feel that the fact that all the $3\times3$ minors are nonpositive (following the same pattern as the third leading principal minor) might be relevant to the fact that the product of the three nonzero eigenvalues being negative, but I can't make the exact connection.
