Timeline for Matrices: system that is "computationally singular" versus "exactly singular" [closed]
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Oct 26, 2018 at 12:42 | history | closed |
Robert Long kjetil b halvorsen♦ Peter Flom |
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| Oct 26, 2018 at 3:05 | review | Close votes | |||
| Oct 26, 2018 at 12:42 | |||||
| Oct 13, 2018 at 23:51 | comment | added | jbowman | Consider an analogous operation: $y = x - z$. If $x=1.0000000123$ and $z = 1.0$, you'll lose 8 digits of accuracy due to subtraction. If $x$ is only accurate to 7 digits, i.e., that "$123$" is just noise, the result is noise, even though the subtraction algorithm is completely accurate. | |
| Oct 13, 2018 at 23:47 | comment | added | jbowman | R is using double-precision arithmetic. It's not an estimation method, it's a calculation. If you want to find the inverse by hand, you can do so (without a calculator) to as many digits of accuracy as you please, at least from a computational standpoint. However, your accuracy is still limited by the accuracy of the values of your matrix; it will do no good to use 32 digit accuracy calculations and leave yourself with 15 digits of accuracy after you've lost 17 digits if the inputs themselves are only accurate to, say, 10 digits, as you will have lost all 10 digits due to ill-conditioning. | |
| Oct 12, 2018 at 1:17 | comment | added | sebelly | Right, but is R using different estimation methods than if I were try and find the inverse of a matrix by hand? | |
| Oct 11, 2018 at 16:51 | comment | added | whuber♦ | Your computer distinguishes small floating point values, like 1.5e-17, from true zeros. One of the differences is that it is capable of dividing numbers by the former but not the latter. | |
| Oct 11, 2018 at 8:09 | history | edited | Ferdi | CC BY-SA 4.0 |
deleted 7 characters in body
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| Oct 11, 2018 at 8:05 | comment | added | user2974951 | Usually a result of highly correlated variables, you may need to drop some variables. | |
| Oct 11, 2018 at 5:41 | comment | added | user158565 | Sorry, I know very little about R, so I do not know what solve.default() does. My previous comment was based on my guess that solve.default() involves the inverse of square matrix. | |
| Oct 11, 2018 at 3:38 | comment | added | sebelly | Exactly, but I’m wondering why exactly that happens. Why does the first matrix return practically zero and the second return pure zero? | |
| Oct 11, 2018 at 3:11 | answer | added | jbowman | timeline score: 9 | |
| Oct 11, 2018 at 0:19 | comment | added | user158565 | I think it mean 1.58603e-17 is close to zero such that it is hard to perform next calculation, and 0 is zero. | |
| Oct 10, 2018 at 23:10 | review | Close votes | |||
| Oct 11, 2018 at 8:09 | |||||
| Oct 10, 2018 at 22:55 | review | First posts | |||
| Oct 11, 2018 at 3:07 | |||||
| Oct 10, 2018 at 22:51 | history | asked | sebelly | CC BY-SA 4.0 |