Timeline for prove spectral equivalence bounds for inverse fractional power of matrices
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 27, 2022 at 21:52 | comment | added | Iosif Pinelis | @Luna947 : you are welcome. | |
| Jun 27, 2022 at 19:50 | comment | added | Luna947 | Thank you very much for the fruitful information that you provided to me. It was really interesting learning about Loewner's theorem and a great help for closing some remaining gaps in my thesis! | |
| Jun 27, 2022 at 18:55 | comment | added | Iosif Pinelis | @Luna947 : now this is correct. | |
| Jun 27, 2022 at 17:46 | comment | added | Luna947 | sorry maybe i formulated the constant to sloppy. We know $x^\top A x \le c^+ x^\top D x$ and $x^\top A^{-1} x \le \frac{1}{c^-} x^\top D^{-1} x$ holds. Doesn't the addition of both inequalities yield $x^\top (A+A^{-1}) x = x^\top A x + x^\top A^{-1} x \le c^+ x^\top D x + \frac{1}{c^-} x^\top D^{-1} x \le \max\{c^+, 1/c^-\} x^\top (D + D^{-1}) x$ ? | |
| Jun 27, 2022 at 16:28 | vote | accept | Luna947 | ||
| Jun 27, 2022 at 14:17 | comment | added | Iosif Pinelis | @Luna947 : no, this is impossible. | |
| Jun 27, 2022 at 10:41 | comment | added | Luna947 | And let me come up with another question - i think that should be the last one! Is it possible to make a statement concerning the spectral equivalence estimates of the sum, e.g., something like $(A+A^{-1}) \le \max\{c^+, \frac{1}{c^+}\} (D + D^{-1})$ based on the bounds that we deduced so far?? | |
| Jun 27, 2022 at 8:55 | comment | added | Luna947 | ok, so I still have one question: in our case, $A$ and $B$ have eigenvalues in (0,\infty ). so the eigenvalues of $-A$ and $-B$ are in (-\infty ,0) and for $-A^{-1}$ and $-B^{-1}$ they are in in (-\infty ,0) and (-\infty ,0). How can I now deduce the statement using Proposition 2.2 since there it is neccessary to have the eigenvalues in the range $(-1,0)$ ? | |
| Jun 27, 2022 at 1:11 | comment | added | Iosif Pinelis | @Luna947 : You should apply Proposition 2.2 with $A,B$ replaced by $-A,-B$, or by $-A^{-1},-B^{-1}$. In other words, if $0<A\le B$ (in the Loewner sense), then $A^{-1}\ge B^{-1}$ -- which is what one should use here. | |
| Jun 27, 2022 at 0:09 | comment | added | Luna947 | Thank you for the hint. I have seen that remark, but in proposition 2.2 (to which they refer in that remark) they state that in that case, the eigenvalues of both A and D have to be in (-1,0). In our case, due to positive definiteness of A and D, this does obviously not hold. So the statement does not hold? | |
| Jun 26, 2022 at 23:47 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |