Singular values of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices

This question is a generalisation of Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are rectangular matrices which itself is a generalisation of Eigenvalues of $AB$ and $BA$ where $A$ and $B$ are square matrices.

Let $A$ be an $m \times n$ matrix and $B$ an $n \times m$ matrix. Obviously, the matrix products $AB$ and $BA$ are possible. Assume $n \leq m$, such that $AB$ is a weakly larger matrix than $BA$.

Facts:

1. The rank of both $AB$ and $BA$ is at most $n$ (link 1)
2. The number of non-zero eigenvalues of both $AB$ and $BA$ is at most $n$ (link 2)
3. If the eigenvalues of $AB$ are $\lambda_1, \ldots, \lambda_n$, the eigenvalues of $BA$ are also $\lambda_1, \ldots, \lambda_n$ (link 3).

Questions:

1. If the singular values of $AB$ are $\sigma_1, \ldots, \sigma_n$, what can be said about the singular values of $BA$?
2. What does Fact 3, compared with the answer to Question 1, say about the differences and the similarities between eigenvalues and singular values?

There is almost no relationship. For example, if we take $$A = \begin{bmatrix}x & 1 \\ 0 & 0\end{bmatrix}, \quad B = \begin{bmatrix}0 & 0 \\ 1 & y \end{bmatrix}$$ then the singular values of $AB$ are the square roots of the eigenvalues of $$(AB)^{\mathsf T} AB = \begin{bmatrix}1 & y \\ y & y^2\end{bmatrix}$$ so they are $\sqrt{1+y^2}$ and $0$. Similarly, the singular values of $BA$ are $\sqrt{1+x^2}$ and $0$. Even in this simple example, the nonzero singular value in one case can vary pretty much independently of the other case. (They must both be at least $1$, but we can tweak that by changing the $1$ in the matrices to some small $\epsilon>0$.)

By taking determinants, we can conclude that the product of the singular values of $AB$ is $\det(AB)$, while the product of the singular values of $BA$ is $\det(BA)$. So if $A$ and $B$ are both square matrices, the singular matrices in both cases have an equal product $\det(A)\det(B)$, which is some amount of dependence.

On the other hand, by taking tensor products of the construction above, we can start with $2n \times 2n$ square matrices $A$ and $B$ where

• $AB$ has singular values $\sigma_1, \sigma_2, \dots, \sigma_n, 0, 0, \dots, 0$,
• $BA$ has singular values $\sigma'_1, \sigma'_2, \dots, \sigma'_n, 0, 0, \dots, 0$,
• and these are free to vary independently of each other.