3
$\begingroup$

Let $A,B\in M_n(\mathbb{C})$ s.t. $A^3+B^3=0,AB-B^2A^2=I_n$.

According to

Prove $BA-A^2B^2=I$

one can deduce that $A,B$ are invertible and $BA-A^2B^2=I_n$.

Now, when $n=2$, one has: (*) $A^6=I_2$ and $AB-BA$ is nilpotent.

I prove (*) by a PC computation using the Grobner basis theory.

Question 1. Prove (*) with hand.

Question 2. Does (*) remain true when $n>2$ ?

$\endgroup$
0
$\begingroup$

Here is a proof when $n=2$. The case $n>2$ seems to be much more difficult.

Case 1. $AB=BA$. Alors $A^6=I$ in $M_n(\mathbb{C})$. Indeed $(AB)^2-(AB)+I=0$ implies that $AB$ is similar to $diag(-jI_p,-j^2I_q)$ where $j=\exp(2i\pi/3)$ and $p+q=n$; then $(AB)^3=A^3B^3=-I$ and $A^6=I$.

Case 2. $AB\not= BA$. Necessarily, the system $\{I,A,B\}$ is free. Let $A^2=aI+\alpha A,B^2=bI+\beta B$. Then $AB-BA=B^2A^2-A^2B^2=\alpha\beta(BA-AB)$ and $\alpha\beta=-1$. Moreover, $A^3+B^3=(\alpha a+\beta b)I+(a+\alpha^2)A+(b+\beta^2)B=0$ implies $\alpha a+\beta b=a+\alpha^2=b+\beta^2=0$.

We deduce that $\alpha^6=1,\beta=-1/\alpha$ and finally $A^3=-\alpha^3 I$.

Now we show that $trace((AB-BA)^2)=0$.

$BABA=B^3A^3+BA$ implies $(BA)^2-(BA)+I=0$; thus $BA$ is diagonalizable and $spectrum(BA)=\{-j,-j^2\}$ (because $AB\not= BA$); consequently, $spectrum(AB)=\{-j,-j^2\}$.

Finally $tr((AB-BA)^2)=2tr((AB)^2)-2tr(B^2A^2)=$

$2tr((AB)^2)-2tr(AB-I)=2(j+j^4)-2(-j-1-j^2-1)=0$.

Conclusion. $AB-BA$ is nilpotent and $A,B$ are simultaneously triangularizable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.