I don't know if the following is correct ($^tA$ denotes the transpose of $A$).
Let $A \in M_n(\mathbb{R})$, and consider $S = A^tA$. Let $(\lambda_i)_{1 \leq i \leq k}$ be the eigenvalues of $A$, and denote $K = \{ \lambda_i^2 \mid 1 \leq i \leq k\}$. Now if $L$ is the spectrum of $S$ then is it true that $ L = K \cup \{1\} \text{ or } L = K$?
I think I have a proof of this fact but I don't know if this is correct.
So far here are my thoughts :
Let's say $v$ is an eigenvector of $S$ but not an eigenvector of $A$ associated with the eigenvalue $\lambda$. So we have : $A^tAv = \lambda v$. Since $v$ is not an eigenvector of $A$ then the vector $a :=^tAv$ is not colinear to the vector $v$. So it means that the matrix $A$ sends the vector $a$ to $\lambda v$.
Hence if $O$ is the rotation matrix that sends $v$ to $a$ then there exist a matrix $B$ such that $Ba = a$ and $^tA = BO$. Thus since the tranpose of a rotaion matrix is it's inverse we have : $A = O^{-1}(^tB)$. Since $Ba =a$ it means $^tBa = a$ thus : $Aa = O^{-1}a = v$ since $O$ is the rotation matrix that sends $v$ to $a$ then it's inverse is the matrix that sends $a$ to $v$. So we have $\lambda = 1$. This means that if $a$ is not colinear to $v$ and $v$ is an eigenvetor of $S$ then the eigenvalue of $v$ is necessarily $1$. $ \square$
Is what I said correct ?
Thank you !