Given SVD(A) and SVD(B) and B is a diagonal matrix, is there a way or method to construct SVD(AB) ?
- Anybody can ask a question
- Anybody can answer
- The best answers are voted up and rise to the top
There really isn't a simple relationship between the SVD of a product and the SVD of the individual factors.
However, there are methods for forming the SVD of a product of two or more matrices, without forming the matrix product itself (which can be a source of inaccuracy); see for instance this paper by Golub, Solna, and van Dooren.
First of all the QR decomposition of a diagonal matrix B is IB where I is the identity and B is already upper triangular since it's diagonal. Also an upper triangular times a diagonal is upper triangular and not diagonal. The derivation of SVD of BA does not appear to be correct. Unfortunately BU in BUSV* is not unitary. Perhaps the derivation above needs to be corrected?
This won't be a complete answer but maybe a step in the right direction.
Consider the equivalent problem of the SVD of $BA$, where $B$ is diagonal. This is without loss of generality because if we know the SVD of $BA$, then we know the SVD of $(AB)^T=A^TB^T$, which is of the form you mention. We will assume that $B$ has non-zero diagonal elements, such that it has full rank.
Let the SVD of A be $USV^*$, where $U$ and $V$ are unitary, $S$ non-negative diagonal, and $V^*$ is the conjugate transpose of $V$. Further, let $B=QR$ be the unique QR decomposition of $B$ obtained by the Gram-Schmidt method such that the diagonal elements of R are positive, where $Q$ is unitary and $R$ is upper triangular. The positivity of the diagonal elements of $R$ will be useful in what follows.
Since $R$ is upper triangular, and $S$ is diagonal, $RS$ will also be diagonal. Namely, $RS$ will be a diagonal matrix formed of the products of the diagonal elements of $R$ with the corresponding elements of $S$. Since $R$ and $S$ both have positive diagonal elements, so does their product, which we will call $D$. Thus an SVD of $BA$ is given by: