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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. $|B|_1 < \lambda$

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  • $\begingroup$ 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 $B$. $\endgroup$ – LinAlg Oct 15 '18 at 1:07
  • $\begingroup$ @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$ $\endgroup$ – Dushyant Sahoo Oct 15 '18 at 5:38
  • $\begingroup$ How is that too trivial if it leads to the same objective value as a data-dependent $\Lambda$? How should the data affect $\Lambda$? $\endgroup$ – LinAlg Oct 15 '18 at 12:19
  • $\begingroup$ 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. $\endgroup$ – Dushyant Sahoo Oct 15 '18 at 18:13
  • $\begingroup$ There is always an optimum for which $\Lambda=I$, since you can always rescale the columns of $B$. $\endgroup$ – LinAlg Oct 15 '18 at 18:15

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