On Fri, 10 Jan 2020, Allin Cottrell wrote:
3) I'm not so sure about Sven's suggestion that (if I'm
understanding him
right) sqrt(M), for M a real matrix with some negative elements, should
return an appropriate complex matrix. It's clearly defensible, but it might
produce surprising/puzzling results in some cases so I think it requires
further discussion.
This is what R does:
<R>
sqrt(as.matrix(c(1, -2)))
[,1]
[1,] 1
[2,] NaN
</R>
and this is what Octave (therefore, I presume, Matlab) does:
<octave>
> sqrt([1 -2])
ans =
1.00000 + 0.00000i 0.00000 + 1.41421i
</octave>
Personally, I'd stick with the R approach.
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Riccardo (Jack) Lucchetti
Dipartimento di Scienze Economiche e Sociali (DiSES)
Università Politecnica delle Marche
(formerly known as Università di Ancona)
r.lucchetti(a)univpm.it
http://www2.econ.univpm.it/servizi/hpp/lucchetti
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