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I want to decompose a diagonal matrix $\Lambda \in R^{n \times n}$ such that $$ \Lambda \approx A\Sigma A^T $$ where $\Sigma \in R^{k \times k}$ is a diagonal matrix and $A \in R^{n \times k}$ is a dense matrix (or it is non-diagonal matrix), and $k < n$. Is there any decomposition which can be used to solve the above problem?

Edit: Here the approximation implies that $\|\Lambda - A\Sigma A^T\|_F^2$ should be as low as possible.

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  • $\begingroup$ I think it is impossible as we have orthogonal columns... In a slightly circular way of dealing with this we can get the truncated SVD of the matrix $\Lambda \approx U\Sigma V^T$ and that will allow us to select the $\Sigma$ such that it include the maximal absolute values of our $\Lambda$ and $A$ (i.e. $U$) will be just encode the row and the sign of the of the values of $\Sigma$... $\endgroup$ – usεr11852 Nov 14 '18 at 0:11
  • $\begingroup$ @usεr11852 If I do SVD, then I will get diagonal $A$. $\endgroup$ – Dushyant Sahoo Nov 14 '18 at 1:59
  • $\begingroup$ It won't be necessarily diagonal. Just it will have a single non-zero value per row. But, OK, we are just playing with words now. :) $\endgroup$ – usεr11852 Nov 14 '18 at 21:34
  • $\begingroup$ Re the edit: as you surely know, the solution to the less restricted problem of approximating $\Lambda$ as $U\Sigma V^\prime$ is given by the SVD. In this case the SVD yields $\Sigma=\Lambda$ (up to permutation), so all you need to do is remove the smallest $n-k$ singular values, which can be written in the form you require, whence the SVD does the trick. $\endgroup$ – whuber May 19 at 14:39
  • $\begingroup$ @whuber SVD won't give the solution where $A$ is a dense matrix $\endgroup$ – Dushyant Sahoo May 19 at 15:21
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For the diagonal matrix $\Lambda$, its rank is the number of non-zero diagonal elements. If $\DeclareMathOperator{\rank}{\text{rank}} \rank(\Lambda)>k$ then equality is impossible since $$ \rank(A\Sigma A^T)\le \min\{ \rank A, \rank\Sigma , \rank A^T \} \le k $$ so $$ \rank\{ A\Sigma A^T \} < \rank \Lambda. $$ But you didn't ask about equality, but used $\approx$ symbol without explaining in what sense there should be approximation. So then the question is unclear, but at least equality is impossible and any approximation would loose rank.

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  • $\begingroup$ Yes you are right, there won't be any equality that is why I used approximation. I updated the question to make everything more clear. $\endgroup$ – Dushyant Sahoo May 19 at 8:04

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