Symbolic matrices and "integrity" of their inverse

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I have to solve the following problem:

Does a matrix $G\in GL(n,\mathbb{Z})$ exists such that $$ G\times A\times G^{-1}=B $$ being $A,B$ given matrices in $\mathbb{Q}$?

Doing everything by hand, I finally find myself with a bunch of symbolic matrices. Now I have to check if they can lay inside $GL(n,\mathbb{Z})$, i.e. if there are integer values for the variables in the matrix such that the matrix is integer, invertible and with integer inverse.

E.g.: $$\left(\begin{array}{cc}x & 0 \\ 0 & y\end{array}\right)$$ does the trick only for $x=y=1$.

Is there a quick method within Sage to solve that last problem?

Thanks!

asked Oct 17 '11

Jesustc gravatar image Jesustc flag of Germany
190 2 15

updated Oct 18 '11

An integer matrix is in GL(n,Z) (i.e., invertible with integer inverse) if and only if its determinant is 1 or -1 (from wikipedia: a square matrix over a commutative ring is invertible if and only if its determinant is a unit in that ring.) This might help you...

parzan (Oct 18 '11)

It does help me indeed, thanks! I should have known that...

My main concern now is how to find out whether a polynomial equation (det=+/-1) has integer roots or not. "assume()" does not the trick :(

Jesustc (Oct 18 '11)

This is in general an undecidable question - see http://en.wikipedia.org/wiki/Hilbert%27s_tenth_problem. However, for specific cases you might be able to find a solution.

parzan (Oct 18 '11)

"Integrality" would be a better word than "integrity".

John Palmieri (Oct 18 '11)

Thanks a lot for the comments, Parzan and Palmieri! What do you think about erasing the question? I don't think it is useful for anybody.

Jesustc (Oct 28 '11)

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Asked: Oct 17 '11

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Last updated: Oct 18 '11

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