Timeline for R lm() solves a singular system without error
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Feb 19, 2021 at 7:23 | comment | added | F. Privé | Yes, this is the solution I came up with (testing if residuals are almost 0). The problem is that you get an R2 of 50% when the outcome has no variation, that can be very misleading. | |
| Feb 18, 2021 at 14:20 | comment | added | whuber♦ |
You are make much out of nothing. (1) Inspect zapsmall(coefficients(lm(x ~ covar))). (2) Type summary(lm(x ~ covar)). Together these should fully answer your question.
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| Feb 18, 2021 at 14:14 | answer | added | Christoph Hanck | timeline score: 1 | |
| Feb 18, 2021 at 8:54 | comment | added | F. Privé | @Dave Please see my edit. | |
| Feb 18, 2021 at 8:54 | history | edited | F. Privé | CC BY-SA 4.0 |
added 131 characters in body
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| Feb 18, 2021 at 8:37 | comment | added | Dave | What matrix would you expect to be singular, covar$^T$covar? | |
| Feb 18, 2021 at 8:30 | history | asked | F. Privé | CC BY-SA 4.0 |