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If two matrices from GL(2,Z) have the same determinant, will there always be a matrix from SL(2,Z) which transforms one matrix to the other?

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    @bghngh121: By "transform", do you mean "conjugate"? That is, does "$M$ transforms $A$ to $B$" mean $B=M^{-1}AM$?2011-02-18
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    Not conjugate- if matrices A and B from GL(2,Z) have equal determinant, will there always be a matrix C from SL(2,Z) such that A = C B ?2011-02-18
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    Upon reading this question, my interpretation of "$M$ transforms $A$ to $B$" was "$MA = B$". Under this interpretation, the answer is trivially **yes**: take $M = $...2011-02-18
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    Ah, how silly of me.. Thanks Pete.2011-02-18
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    I don't think it will be always in $SL(2,\mathbb{Z})$. Maybe just $SL(2,\mathbb{Q})$.2011-02-19
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    @p791i78: Since $A$ and $B$ are both invertible over $\mathbb{Z}$ the product $BA^{-1}$ lies in $\mathrm{GL}(2,\mathbb{Z})$; the determinant is $1$ (since $\det(A)=\det(B)$, and they are both either $1$ or $-1$, so $\det(A^{-1}) = \det(A)$), and $(BA^{-1})A = B$. So $C=BA^{-1}$ is the matrix sought, it *always* lies in $\mathrm{SL}(2,\mathbb{Z})$.2011-02-19
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    Please do not use answers to make comments.2011-02-19
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    @Qiaochu: The user does not seem to have enough reputation to leave comments, though. I guess he could go around the site trying to get enough reputation so that he can finally write down his comment on this question, but still...2011-02-19

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