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We have $A,B$ two $3×3$ matrices with integer numbers. We know that $(AB)^{2}+BA=(BA)^2+AB$.

a) Show that $\det((AB)^{n}-(BA)^{n})$ is divisible by $det(AB-BA)$.

b) Show that if $\det(AB-BA)=1$, then $\det((AB)^{n}-(BA)^{n})$ is a perfect cube.

I have tried taking the trace in the first equality but nothing interesting. Maybe will help us rewriting $\det(AB-BA)$ as $Tr((AB-BA)^{3})/3$? It feels like we need to use some induction here but I dont know what is the "induction general form". I see we can rewrite $\det(AB-BA)=\det((AB)^{2}-(BA)^{2})$. I think that perfect cube will come from a determinant rewriting in polynomial form, but I dont know how we can rewrite it as a polynomial.

Sebastiano
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Stefan Solomon
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  • To get the proper font and spacing for $\det$, use \det. For operators that don't have a command of their own, you can use \operatorname{name}. – joriki Jan 09 '23 at 23:43
  • While you have shown effort which is solid, the close votes and downvotes (the combination of which seems a tad harsh to me) likely stems from a combination of this sounding very much like a homework question (for which the site policy is to make it clear that's where the question is from), the TeX could use a little work, and you posted two questions (they're related so I get it, but people will be more critical of the minutia in your post if you have a sequence of questions). – Brevan Ellefsen Jan 09 '23 at 23:45

1 Answers1

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Since $AB$ is an integer matrix, its characteristic polynomial $\det(xI-AB)$ has integer coefficients. Let us write it as $$ x^3-a_3x^2-b_3x-c_3. $$ So, for every $n\ge3$ we have $$ (AB)^n=a_n(AB)^2+b_nAB+c_nI $$ where the case $n=3$ is a consequence of Cayley-Hamilton theorem and when $n>3$, $a_n,b_n$ and $c_n$ are integer coefficients obtained by the following recurrence relations: $$ \begin{aligned} a_n&=a_{n-1}a_3+b_{n-1},\\ b_n&=a_{n-1}b_3+c_{n-1},\\ c_n&=a_{n-1}c_3.\\ \end{aligned} $$ As $AB$ and $BA$ have the same characteristic polynomial (Sylvester's secular theorem), we also have $$ (BA)^n=a_n(BA)^2+b_nBA+c_nI. $$ Let $f(n)=\det\left((AB)^n-(BA)^n\right)$. Then $f(0)=0$ and $f(1)=\det(AB-BA)$. By the given condition, we have $(AB)^2-(BA)^2=AB-BA$, so that $f(2)=\det(AB-BA)$ too. When $n\ge3$, $$ \begin{aligned} f(n)&=\det\left((AB)^n-(BA)^n\right)\\ &=\det\left(a_n\left((AB)^2-(BA)^2\right)+b_n(AB-BA)\right)\\ &=\det\left((a_n+b_n)(AB-BA)\right)\\ &=(a_n+b_n)^3\det(AB-BA).\\ \end{aligned} $$ Hence the results follow.

user1551
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