Is there any relationship between the eigenvalues of $A$ and $AA^T$ I am trying to implement some deep learning algorithm based on the derivatives of eigenvalues. The problem that I have is that this line in Pytorch is not implemented which doesn't allow me to calculate the derivatives of complex eigenvalues for real inputs (which is what I want to do)...
I have no idea on how to implement this or if it is even possible yet, but a quick solution may be to instead look at the eigenvalues of $AA^T$ which are real. If I could find some relationship between the eigenvalues of $A$ and $AA^T$ I may be able to constrain the eigenvalues of $A$ by using already implemented derivatives for the real eigenvalues of $AA^T$.
So is there any relationship between these eigenvalues? I cannot figure out where to search for this kind of information...
I plotted histograms of the two eigenvalues and found very different distributions, but IDK where to go from here. Does anyone know?

 A: The eigenvalues of $AA^T$ are the squares of the singular values of $A$. In general, it is not possible to find the eigenvalues of $A$ using (only) its singular values or the singular values of $A$ using (only) its eigenvalues.
That said, there are several notable results that describe the relationship between singular values (typically denoted $\sigma_i$) and eigenvalues (typically denoted $\lambda_i$), but one powerful result of this kind is the set of inequalities
$$
\prod_{j=1}^k |\lambda_j| \leq \prod_{j=1}^k \sigma_j, \quad 1 \leq k \leq n,
$$
where $n$ is the size of the matrix $A$ and the indices are chosen such that $\sigma_1 \geq \cdots \geq \sigma_n$ and $|\lambda_1| \geq \cdots \geq |\lambda_n|$. In fact, the two products are necessarily equal for $k = n$ (with both sides equal to $|\det(A)|$. This result implies a more general result known as "Weyl's Majorant Theorem", which leads to a lot of inequalities of this kind. For instance, we have
$$
\sum_{j=1}^k |\lambda_j|^p \leq \sum_{j=1}^k \sigma_j^p, \quad 1 \leq k \leq n
$$
for all exponents $p \geq 0$.
In a sense, there is no result that yields a stronger relationship in general than this. In particular, Weyl's Majorant theorem has the following converse: if $\lambda_i, \sigma_i$ are complex and non-negative numbers (respectively) ordered with non-increasing magnitude such that we have
$$
\prod_{j=1}^k |\lambda_j| \leq \prod_{j=1}^k \sigma_j, \quad 1 \leq k \leq n-1,\\
\prod_{j=1}^n |\lambda_j| = \prod_{j=1}^n \sigma_j,
$$
then there necessarily exists a matrix $A$ whose eigenvalues are equal to the $\lambda_i$ and whose singular values are equal to the $\sigma_i$.
