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Let $A \in S^{n}_{+}$ be a positive semi-definite matrix and $D \in S^{n}_{+}$ a diagonal matrix with all the diagonal entries no smaller than one, i.e., $D_{ii} \geq 1$ for all $i \leq n$.

I wonder whether the transformation $DAD$ will scale up the eigenvalues? i.e., let $\lambda_i(M)$ denote the $i$-th largest eigenvalue of a matrix $M$; is it always true that $\lambda_i(DAD) \geq \lambda_i(A) \geq 0$ for all $i \leq n$?

I feel it is intuitive but haven't formally proved/disproved it yet. Forgive me if it is a commonly-known or obvious result.

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  • $\begingroup$ @FedorPetrov. $D$ is diagonal. $\endgroup$ Mar 12, 2021 at 7:29

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Yes. Write the eigenvalues as $\max\min$ (or as $\min\max$) of Rayleigh quotient. Then use the fact that $\|Dx\|\ge\|x\|$.

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