Re: Nth root of a matrix (with some constraints)

In article
<woodchips-E4871B.20261518112005@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>, John
D'Errico <woodchips@xxxxxxxxxxxxxxxx> wrote:

> .............
> But clearly A1 and B are not the same. So there
> may well exist an n'th root that has positive
> elements.
>
> John
-------------------
Hello John,

I claim the following. Let PN be a 3 x 3 matrix having three
eigenvalues with distinct magnitudes and having eigenvectors V such that
PN = V*D*inv(V), as is true for cyclist's problem. Then for any whole
number N, any P1 such that P1^N = PN must have as eigenvalues, some N-th
roots of those of PN which must therefore also have distinct magnitudes,
and furthermore P1 must be expressible as P1 = V*D^(1/N)*inv(V) with those
particular eigenvalue roots in D^(1/N).

This being the case, then there are only N^3 candidates for a valid
probabilistic type solution for each N. This greatly narrows down the
field of possibilities. However, I grant you, it remains theoretically
possible that somewhere among all these cases there might lurk one for
some N that has the necessary properties. However, I wrote a routine that
went through, brute force, all N^3 possible cases for each N from 1 up to
50 for the given PN, allowing a healthy round off tolerance, and no
possibilities popped up. (It took my ancient system over 11 minutes to
get that far, so I quit there.) In view of that fact, prospects for a
solution to cyclist's problem for any N at all still seem to me to be very
bleak.

(Remove "xyzzy" and ".invalid" to send me email.)
Roger Stafford
.

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