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:


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*The rank of both $AB$ and $BA$ is at most $n$ (link 1)

*The number of non-zero eigenvalues of both $AB$ and $BA$ is at most $n$ (link 2)

*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:


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*If the singular values of $AB$ are $\sigma_1, \ldots, \sigma_n$, what can be said about the singular values of $BA$?

*What does Fact 3, compared with the answer to Question 1, say about the differences and the similarities between eigenvalues and singular values?

 A: 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.

