This is a simpler version of this question. Let $x=\left(\begin{array}{lll} 2 & 0 & 0\\\ 0& 1 & 0\\\ 0 & 0 & \frac12\end{array}\right)$. Is there a $3\times 3$-matrix $y$ with complex entries and $\det(y)=1$ such that $[x,y,y]=x$? Here $[a,b]=a^{-1}b^{-1}ab$, $[a,b,c]=[[a,b],c]]$? Perhaps somebody with a good Groebner basis software can check it.
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1 Answer
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Does this do you?
Magma V2.11-11 Sun Jul 24 2011 20:03:50 on sevilla [Seed = 2330466759]
Type ? for help. Type -D to quit.
> R<[x]> := PolynomialRing(Rationals(),9,"grevlex");
> y := Matrix(3,x);
> d := DiagonalMatrix(R,[2,1,1/2]);
> m1 := d^-1*Adjoint(y)*d*y;
> m2 := Adjoint(m1)*Adjoint(y)*m1*y;
> I := ideal<R|Eltseq(m2-d),Determinant(y)-1>;
> Groebner(I);
> I;
Ideal of Polynomial ring of rank 9 over Rational Field
Graded Reverse Lexicographical Order
Variables: x[1], x[2], x[3], x[4], x[5], x[6], x[7], x[8], x[9]
Groebner basis:
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$\begingroup$ Thank you! It is what I expected. Now of course I would like to know why it is so but it is a different question. $\endgroup$– user6976Jul 25, 2011 at 4:18