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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.)

questioner
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    You don't mean $DAD^{-1}$, right? Similar matrices have the same eigenvalues (http://math.stackexchange.com/questions/8339/similar-matrices-have-the-same-eigenvalues-with-the-same-geometric-multiplicity) – parsiad Aug 08 '15 at 03:21
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    Absolutly, I meant $DAD$. You're right about similar matrices though. – questioner Aug 08 '15 at 04:09
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    It's impossible to give an exact result without more information about $A$. However, we might be able to get some useful inequalities. – Ben Grossmann Aug 08 '15 at 04:28
  • 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) – questioner Aug 08 '15 at 04:32
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    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. – Ben Grossmann Aug 08 '15 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 $$\alpha_1 \cdot \beta_1\ge \gamma_1 \\ \alpha_1 \alpha_2 \cdot \beta_1 \beta_2\ge \gamma_1 \gamma_2\\ \ldots\ldots\ldots\\ \alpha_1 \cdots\alpha_n \cdot\beta_1 \cdots \beta_n = \gamma_1 \cdots \gamma_n$$ 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.

luckreez
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orangeskid
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