Timeline for Inverse of a matrix and the inverse of its diagonals
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Mar 22, 2023 at 21:25 | comment | added | Fedor Petrov | @heroxav I am afraid that with fixed $D$ the matrix $R$ may be almost singular which makes $R^{-1}$ too large. Check the matrix $\pmatrix{1&1-\varepsilon\\1-\varepsilon&1}$. | |
| Mar 22, 2023 at 20:14 | comment | added | heroxav | @FedorPetrov Can one also give an upper bound of similar flavor, that is, $x^\top R^{-1} x \le c \cdot x^\top D^{-1} x$? | |
| Mar 7, 2019 at 20:18 | comment | added | Mamal | I got it. Thank you. | |
| Mar 7, 2019 at 20:14 | vote | accept | Mamal | ||
| Mar 7, 2019 at 20:12 | comment | added | Fedor Petrov | We know that $D^{1/2}$ exists, and so does $D^{-1/2}$, and we define $Q$ as $Q:=D^{-1/2}RD^{-1/2}$. | |
| Mar 7, 2019 at 20:10 | comment | added | Fedor Petrov | There is no theory, I simply define $Q$ this way. | |
| Mar 7, 2019 at 20:02 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Mar 7, 2019 at 20:01 | comment | added | Fedor Petrov | $Q=D^{-1/2}RD^{-1/2}$, diagonal elements of $Q$ are equal to 1. | |
| Mar 7, 2019 at 19:57 | history | edited | Fedor Petrov | CC BY-SA 4.0 |
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| Mar 7, 2019 at 19:48 | history | answered | Fedor Petrov | CC BY-SA 4.0 |