Relating the SVD of $AB^T$ and $B^TA$

Question

Let $C$ be a $m \times n$ real matrix. We can show that $CC^T$ and $C^TC$ have the same non zero eigenvalues and their respective eigenvectors can be related through $Cv = \lambda^{-1/2} u$, $C^Tu = \lambda^{-1/2} v$ where $v$ is an eigenvector of $C^TC$ and $u$ of $CC^T$ with same eigenvalue $\lambda > 0$. This is the principle of the singular value decomposition.

I am wondering if we can extend this connection to the following. Let $A,B$ be $m \times n$ matrices, is there a way to relate the singular values and (left/right) singular vectors of the matrices $AB^T$ and $B^TA$?

Answer 1

Related result from Higham's "Functions of Matrices" book

• Harley Flanders. Elementary divisors of AB and BA. Proc. Amer. Math. Soc., 2(6): 871–874, 1951. (Cited on p. 28.