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To reproduce

set.seed(1)
N <- 100
x <- rep(1, N)
covar <- matrix(rnorm(N * 10), N)
lm(x ~ covar)

Because of the intercept, I would expect this to be singular and not solvable. Instead, I just get very small values for the coefficients of covar.

Anyone knows a way around this?

Edit: what is singular is

model.matrix(~ ., data = cbind.data.frame(x, covar))

(first two columns are all 1s)

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  • $\begingroup$ What matrix would you expect to be singular, covar$^T$covar? $\endgroup$
    – Dave
    Commented Feb 18, 2021 at 8:37
  • $\begingroup$ @Dave Please see my edit. $\endgroup$
    – F. Privé
    Commented Feb 18, 2021 at 8:54
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    $\begingroup$ 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. $\endgroup$
    – whuber ♦
    Commented Feb 18, 2021 at 14:20
  • $\begingroup$ 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. $\endgroup$
    – F. Privé
    Commented Feb 19, 2021 at 7:23

1 Answer 1

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I am not sure why this should not be solvable. You regress, as you point out, a constant on a constant, which is "generated" to affect the dependent variable with a coefficient of one, plus other variables which are generated in a way so as to have nothing to do with the dependent variable. So a unit coefficient on the constant term and zero coefficients for the rest seem precisely what to expect. That they are not exactly zero is numerical noise to me.

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