Let $A \in \mathbb{R}^{n \times n}$ be a positive definite matrix and $D = \text{diag} (d_1, d_2, ... , d_n)$ be a positive diagonal matrix. We know that eigenvalues of A are $\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n$. What will be eigenvalues of $DAD$? (Obviously they're all positive.)
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2$\begingroup$ You don't mean $DAD^{-1}$, right? Similar matrices have the same eigenvalues (math.stackexchange.com/questions/8339/…) $\endgroup$– parsiadAug 8, 2015 at 3:21
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1$\begingroup$ Absolutly, I meant $DAD$. You're right about similar matrices though. $\endgroup$– questionerAug 8, 2015 at 4:09
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2$\begingroup$ It's impossible to give an exact result without more information about $A$. However, we might be able to get some useful inequalities. $\endgroup$– Ben GrossmannAug 8, 2015 at 4:28
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$\begingroup$ Assume $A$ is laplacian of an undirected connected graph. (I know $A$ is positive semi-definite in this case). What inequalities? (beside the most obvious ones) $\endgroup$– questionerAug 8, 2015 at 4:32
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1$\begingroup$ I'd have to look through Bhatia's matrix analysis to see if something applies; I can't think of anything relevant off the top of my head. $\endgroup$– Ben GrossmannAug 8, 2015 at 12:37
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You do have in general inequalities for singular values of the product of two matrices $M$, $N$, due to Weyl or Schur, I forgot, Horn and Johnson should have the reference. If $\alpha_i$, $\beta_i$ $\gamma_i$ are the singular values of $M$, $N$ and $MN$, in decreasing order then $$\alpha_1 \cdot \beta_1\ge \gamma_1 \\ \alpha_1 \alpha_2 \ge \gamma_1 \gamma_2\ge \gamma_1 \gamma_2\\ \ldots\ldots\ldots\\ \alpha_1 \cdots\alpha_n \cdot\beta_1 \cdots \beta_n = \gamma_1 \cdots \gamma_n$$ The first inequality is not hard, since it is the inequality for the $l^2$ norm. The other are obtained considering the associated operators $\wedge^k M, N, MN$.
So, fixing $\lambda_i$ for a hermitian, and the $d_i$ there definitely are inequalities for the eigenvalues of $DAD$. Hard to tell what are all the defining inequalities.
answered Dec 28, 2020 at 4:01