Timeline for Tight bounds on maximum and minimum eigenvalues of product of a matrix with a diagonal matrix: of the form $\ A^T D A$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 14, 2018 at 6:53 | comment | added | Samrat Mukhopadhyay | @Robert, see Rayleigh quotient for a derivation. | |
| Jan 13, 2018 at 21:42 | comment | added | Robert | Sorry, but why is it easy to see the inequality bounds? | |
| Dec 7, 2016 at 7:23 | comment | added | Samrat Mukhopadhyay | Oh. Actually the matrices $A$ that I am interested in have more rows than columns, have full column rank and their eigenvalues lie in a range $[1-\delta,1+\delta]$, where $\delta\in (0,1)$. Does that make the situation any better? Actually I am not been able to grasp what will happen if $D$ has entries, for example, all $0$,except only one diagonal element $1$. Is it possible then that the difference between the bound and the actually eigenvalues are very small? | |
| Dec 7, 2016 at 5:52 | comment | added | Robert Israel | If it's very close to $I$, the inequalities will be very close to equalities. There's no hope of improving the estimates without something that specifies how far $A$ is from matrices that make the inequality an equality. | |
| Dec 5, 2016 at 10:17 | history | edited | Samrat Mukhopadhyay | CC BY-SA 3.0 |
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| Dec 5, 2016 at 8:06 | comment | added | Samrat Mukhopadhyay | Yes, I know. But I really want to know about the cases where $A$ is not diagonal, and probably not even square, so that we do not have trivial examples like this. I will include that in my question. Thanks. | |
| Dec 4, 2016 at 21:14 | comment | added | Robert Israel | If $A=I$, your inequalities are equalities. | |
| Dec 4, 2016 at 15:02 | history | edited | Samrat Mukhopadhyay | CC BY-SA 3.0 |
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| Dec 4, 2016 at 14:42 | history | asked | Samrat Mukhopadhyay | CC BY-SA 3.0 |