# Decomposition of a diagonal matrix

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.

• 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 $\Sigma$... – usεr11852 Nov 14 '18 at 0:11
• @usεr11852 If I do SVD, then I will get diagonal $A$. – Dushyant Sahoo Nov 14 '18 at 1:59
• 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. :) – usεr11852 Nov 14 '18 at 21:34
• 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. – whuber May 19 at 14:39
• @whuber SVD won't give the solution where $A$ is a dense matrix – Dushyant Sahoo May 19 at 15:21

## 1 Answer

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.

• 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. – Dushyant Sahoo May 19 at 8:04