TITLE: Singular value decomposition of product of matrices
QUESTION [9 upvotes]: Given SVD(A) and SVD(B) and B is a diagonal matrix, is there a way or method to construct SVD(AB) ?

REPLY [19 votes]: Actually there is an answer, but somewhat complex. 
Denote the SVD decomposition of $A$ by $U^*S^*V'$, denote $\operatorname{qr}(BU) = QR$, and denote SVD decomposition of $RS$ by $U_1^*S_1^*V_1'$ then the SVD decomposition of $BA$ is $(QU_1)S_1(VV_1)'$. 
Proof: $BA= BUSV'=QRSV'=(QU_1)S_1(V_1'V') = (QU_1)S_1(VV_1)'$. $QU_1$ is unitary and $VV_1$ is unitary and $S_1$ is diagonal. 
Interpretation: the singular values of $BA$ are the singular values of $RS$. So the original singular values are multiplied by the upper triangular $R$ of the qr decomposition of $BU$, i.e., $B$ after unitary transformation by right bases $U$ of $A$.