4
$\begingroup$

Is there any way to convert the output of NCAlgebra to standard Mathematica form?

For example 

<< NC`

<< NCAlgebra`

MatrixForm[m = {{a, b, c}, {d, e, f}, {g, h, i}}];
NCInverse[m][[1, 1]]

The output is

    a^(-1) ** (1 - b ** (e - d ** a^(-1) ** b)^(-1) ** (-d ** a^(-1) - (f - d ** a^(-1) ** c) ** (i - g ** a^(-1) ** c - (h - g ** a^(-1) ** b) ** (e - d ** a^(-1) ** b)^(-1) ** (f - d ** a^(-1) ** c))^(-1) ** (-g ** a^(-1) + (h - g ** a^(-1) ** b) ** (e - d ** a^(-1) ** b)^(-1) ** d ** a^(-1))) - c ** (i - g ** a^(-1) ** c - (h - g ** a^(-1) ** b) ** (e - d ** a^(-1) ** b)^(-1) ** (f - d ** a^(-1) ** c))^(-1) ** (-g ** a^(-1) + (h - g ** a^(-1) ** b) ** (e - d ** a^(-1) ** b)^(-1) ** d ** a^(-1)))

I want to change the output to standard form of Mathematica like

Inverse[a].(1-b.Inverse[e-d.Inverse[a].b]...
Improve this question
$\endgroup$
1
  • $\begingroup$ This is tricky, especially if you want it to work with letters that will be substituted by matrices. The main trouble is that things like a + b ** c will be expanded incorrectly: b**c will be added entrywise to the entries a. We have an internal routine for doing that in NCAlgebra but I never thought it was good enough to be made public. $\endgroup$ Commented Dec 11, 2018 at 6:54

1 Answer 1

Reset to default
6
$\begingroup$

While I am not familiar with NCAlgebra, the following seems to work:

result /. {inv -> Inverse, NonCommutativeMultiply -> Dot}
(* Inverse[
  a].(1 - b.Inverse[
     e - d.Inverse[
        a].b].(-d.Inverse[a] - (f - d.Inverse[a].c).Inverse[
        i - g.Inverse[a].c - (h - g.Inverse[a].b).Inverse[
           e - d.Inverse[a].b].(f - d.Inverse[a].c)].(-g.Inverse[
            a] + (h - g.Inverse[a].b).Inverse[
           e - d.Inverse[a].b].d.Inverse[a])) - 
   c.Inverse[
     i - g.Inverse[a].c - (h - g.Inverse[a].b).Inverse[
        e - d.Inverse[a].b].(f - d.Inverse[a].c)].(-g.Inverse[
         a] + (h - g.Inverse[a].b).Inverse[
        e - d.Inverse[a].b].d.Inverse[a])) *)

Note that NCAlgebra produces inv[a] when inverting a. This is formatted as $a^{-1}$, but the underlying structure is just inv. You can see this using FullForm.

Improve this answer
$\endgroup$

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.