Timeline for Asymptotic expansion involving a matrix equation
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jul 23, 2020 at 10:35 | vote | accept | Rajesh D | ||
| Jul 19, 2020 at 8:55 | comment | added | Rajesh D | Thanks for your valuable comments and answer. | |
| Jul 19, 2020 at 8:51 | comment | added | Federico Poloni | @RajeshDachiraju That I don't know. | |
| Jul 19, 2020 at 8:43 | comment | added | Federico Poloni | @RajeshDachiraju Still won't work. What if $C$ is entrywise $O(\lambda^{-1})$ but has an eigenvalue that goes like $O(\lambda^{-3})$? For instance, take $n>2$ and let $Q$ be an orthogonal matrix such that $Ef=f_1$, where $f$ is the vector of all ones and $f_1$ is the first column of $I$. Then $QAQ^{-1} = diag(n,0,0,...0)$. Take then $QCQ^{-1} = diag(0,\lambda^{-1},\lambda^{-3},\lambda^{-3},\dots,\lambda^{-3})$. Then if $Q$ is "complicated enough" all entries of $Q$ will scale like $\lambda^{-1}$, but $QMQ^{-1} = diag(n+\lambda^{-2},\lambda^{-1},\lambda^{-3},\lambda^{-3},\dots,\lambda^{-3})$. | |
| Jul 19, 2020 at 8:37 | comment | added | Rajesh D | Your counter example seems valid to me. Luckily in my problem $C$ goes as $\frac{1}{\lambda}$...perhaps $O$ is not suitable here, I should have used $\omega$ or something. I have a closed form expression for $C$ which is $\frac{1}{\lambda}$. Thank you for the answer. | |
| Jul 19, 2020 at 8:35 | history | edited | Federico Poloni | CC BY-SA 4.0 |
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| Jul 19, 2020 at 8:34 | comment | added | Federico Poloni | EDIT: Ok, fixed counterexample, hopefully. | |
| Jul 19, 2020 at 8:30 | history | edited | Federico Poloni | CC BY-SA 4.0 |
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| Jul 19, 2020 at 8:25 | comment | added | Rajesh D | $M$ is positive definite. | |
| Jul 19, 2020 at 8:25 | comment | added | Rajesh D | Your example still assumes $\rho(M) = 0$ which is not possible. | |
| Jul 19, 2020 at 8:23 | comment | added | Federico Poloni | @RajeshDachiraju Updated the counterexample. | |
| Jul 19, 2020 at 8:22 | history | edited | Federico Poloni | CC BY-SA 4.0 |
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| Jul 19, 2020 at 8:22 | comment | added | Rajesh D | In question I had stated $A+C$ is known to be $psd$ so $M$ is a pd. Hope that solves the problem. | |
| Jul 19, 2020 at 8:21 | comment | added | Rajesh D | $M$ is positive definite, as $A+C$ is psd. | |
| Jul 19, 2020 at 8:19 | history | answered | Federico Poloni | CC BY-SA 4.0 |