Matrix logarithm

[GUSTAVO TERRA BASTOS]

Hi guys. 

Given two n x n matrices M, N, we know it is a big problem to find the positive integer "i" such that M^i = N (There are other hypothesis involved). In my particular case, I would like to do the same for 3 x 3 matrices M , N over F_{11^2} (finite field with 121 elements). 

Is it possible to do in a regular PC ?

Best regards!

Gustavo

[David Joyner]

On Mon, Apr 18, 2022 at 9:31 AM GUSTAVO TERRA BASTOS
<gtba...@ufsj.edu.br> wrote:
>
> Hi guys.
>
> Given two n x n matrices M, N, we know it is a big problem to find the positive integer "i" such that M^i = N (There are other hypothesis involved). In my particular case, I would like to do the same for 3 x 3 matrices M , N over F_{11^2} (finite field with 121 elements).
>

FYI, some options are mentioned in
https://math.stackexchange.com/questions/3116315/is-there-a-matrix-logarithm-in-sage

> Is it possible to do in a regular PC ?
>
> Best regards!
> Gustavo
>

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[Oscar Benjamin]

On Mon, 18 Apr 2022 at 14:30, GUSTAVO TERRA BASTOS <gtba...@ufsj.edu.br> wrote:
>
> Hi guys.
>
> Given two n x n matrices M, N, we know it is a big problem to find the positive integer "i" such that M^i = N (There are other hypothesis involved). In my particular case, I would like to do the same for 3 x 3 matrices M , N over F_{11^2} (finite field with 121 elements).

Note that finding this integer i here is something different from what
is usually referred to as the "matrix logarithm" (the inverse of the
matrix exponential) which gives a matrix rather than an integer as the
result.

Is it prohibitive to just compute powers of M until you find that M^i = N?

Any i would need to satisfy det(M)^i = det(N) which should be faster
to solve for large i unless det(M) and det(N) are both zero or one.
Otherwise I think you can generalise the approach to other
coefficients in the characteristic polynomial.

Oscar