# 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? I have asked the same question here https://stats.stackexchange.com/questions/376822/decomposition-of-a-diagonal-matrix

Edit: The number of zero elements in each row and column of $$A$$ matrix should not be greater than half the size of the row and column respectively.

For simplicity of explanation, suppose that $$\Lambda = \pmatrix{\lambda_1 \\ & \lambda_2 \\ && \ddots \\ &&& \lambda_n}$$ with $$|\lambda_1| \geq |\lambda_2| \geq \cdots \geq |\lambda_n|$$. The Eckart-Young theorem tells us that the best rank-$$k$$ approximation (or at least one of the best rank-$$k$$ approximations) of $$\Lambda$$ (relative to the Frobenius norm or induced $$2$$-norm) will be the diagonal matrix $$\Lambda^{(k)} = \pmatrix{\lambda_1 \\ & \ddots \\ && \lambda_k \\&&&0\\&&&&\ddots\\ &&&&&0}$$ We can write $$\Lambda^{(k)} = M DM^T$$, where $$M = \pmatrix{\sqrt{|\lambda_1|}\\ & \ddots \\ && \sqrt{|\lambda_k|}\\&\mathbf 0_{(n-k)\times k}}$$ and $$D$$ is a $$k\times k$$ diagonal matrix with $$\pm 1$$ on the diagonal. If $$U$$ is an orthogonal matrix that satisfies $$UD = DU$$, then we have $$(MU)D(MU)^T = MUDU^TM^T = MDUU^TM = MDM^T = \Lambda^{(k)}$$ I suspect that selecting a dense orthogonal matrix satisfying $$UD = DU$$ will give you a good result.
Regarding orthogonal matrices that commute with $$D$$: if $$D$$ is such that the $$1$$s and $$-1$$s are grouped together, then we have $$D = \pmatrix{I_{p \times p} \\ & - I_{(n-p)\times (n-p)}}$$ An orthogonal matrix $$U$$ will satisfy $$UD = DU$$ if and only if it is conformably block-diagonal, i.e. if and only if we have $$U = \pmatrix{U_{p \times p}^{(1)} & 0\\ 0 & U_{(n-p) \times (n-p)}^{(2)}}$$ and notably, the above $$U$$ will be orthogonal if and only if both $$U^{(1)}$$ and $$U^{(2)}$$ are orthogonal.
• Thanks for the answer, but the $MU$ matrix has $n-k$ columns and rows completely 0 equal to zero. I want the matrix to be dense, and also it is not necessary to have minimum Frobenius norm. The main condition is that $A\Sigma A^T$ should be closed to diagonal matrix and $A$ is dense. – Dushyant Sahoo Nov 15 '18 at 6:43