Timeline for All and the only algebraically closed fields s.t. any regular n-by-n matrix has a k-th root for every k
Current License: CC BY-SA 3.0
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| Mar 17, 2013 at 21:18 | comment | added | Salvo Tringali | I had a look at Higham's Functions of Matrices - Theory and Applications, before asking. And even if the 7th chapter of the book is entirely focused on $k$-th roots of complex matrices, nothing like your (absolutely brilliant) proof seems to be there. And the 3rd edition of the classical Matrix Computations by Golub and Van Loan spends no word for arbitrary $k$-th roots, while square roots are just mentioned in relation to the Cholesky and polar decompositions (and again, there's nothing in the lines of your slick argument). | |
| Mar 17, 2013 at 20:08 | comment | added | Peter Mueller | I don't know a reference, but I'm pretty sure that it should be at many places. Maybe mostly only over the complex numbers, but that doesn't make a difference. All we need is that the binomial coefficients $\binom{1/k}{i}$ exist in the base field. | |
| Mar 17, 2013 at 20:01 | comment | added | Salvo Tringali | Wonderful! Do you know of any reference to (the 2nd part of) your proof (other than the link to this answer)? | |
| Mar 17, 2013 at 20:01 | vote | accept | Salvo Tringali | ||
| Mar 17, 2013 at 19:45 | history | answered | Peter Mueller | CC BY-SA 3.0 |