# Constraint on product of matrices

I am trying to solve the below optimization problem

\begin{equation*} \begin{aligned} & \underset{{A}, {B}, {\Lambda}}{\text{minimize }} & & \|X - AB\Lambda C^TD^T|_F^2 \\ & \text{subject to} & & \Lambda \text{ and } B\Lambda C^T \text{ are diagonal matrices}\\ \end{aligned} \end{equation*} where $$X \in R^{n\times n}, A\in R^{n\times k_1}, D\in R^{n\times k_1}, B \in R^{k_1\times k_2}, C \in R^{k_1\times k_2}, \Lambda \in R^{k_2\times k_2}$$, $$X$$ is known and $$n > k_1 > k_2$$. I am not sure how to put a constraint on $$B$$, $$C$$ and $$\Lambda$$ such that $$B\Lambda C^T$$ is a diagonal matrix. I know I can solve for all unknown using alternative minimization but I am not sure how to include the constraint. I have asked a similar question here Diagonal constraint on product of matrices

Edit 1: $$A, B, C$$ and $$D$$ are sparse matrices i.e. $$|B|_1 < \lambda$$

• You can choose $\Lambda=I$ without losing any freedom. You then get linear constraints on $B$ from the requirement that $BC^T$ is diagonal. The difficulty is in the product of $A$ and $B$. – LinAlg Oct 15 '18 at 1:07
• @LinAlg Is it possible to get a method completely data-driven? It will be too trivial if I choose $\Lambda = I$, $\Lambda = I$ is one of the solutions but I don't want to make any assumptions about \Lambda$– Dushyant Sahoo Oct 15 '18 at 5:38 • How is that too trivial if it leads to the same objective value as a data-dependent$\Lambda$? How should the data affect$\Lambda$? – LinAlg Oct 15 '18 at 12:19 • The optimization problem is non-convex, so it should have multiple local minima and$\Lambda = I$might lead to some of the local minima. Are you saying for all the solutions$\Lambda = I$? I think I didn't understand why are you saying that we can choose$\Lambda = I$without losing any freedom. – Dushyant Sahoo Oct 15 '18 at 18:13 • There is always an optimum for which$\Lambda=I$, since you can always rescale the columns of$B\$. – LinAlg Oct 15 '18 at 18:15