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Let $A,B \in \mathcal{M}_{n}(\mathbb{R})$ such that $$A^2+B^2=\sqrt{2+\sqrt{2}}\cdot AB$$ Knowing that $\det(AB-BA)>0$, prove that $n$ is multiple of $16$.

I know that for this type of problems, one usually uses some identities such as $$(A+iB)(A-iB)=c(AB-BA)$$ and its conjugate, where $c$ is a complex number and since $A,B$ are real matrices, their determinants will be positive real numbers. Since $\det(AB-BA)>0$, that will lead to $c^n=0$ and with the help of some trigonometry it would follow that $n$ is the multiple of something.

Here, however, I couldn't obatain the identity I described. That square root points, though, exactly to this method and some trigonometry.

AndrewC
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1 Answers1

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We show that if $A,B\in M_n(\mathbb{C})$ satisfy the above equation and are s.t. $\det(AB-BA)\not= 0$, then $n=0 \;mod(8)$.

Proof. Let $\alpha=\sqrt{2+\sqrt{2}},\beta=\sqrt{2-\sqrt{2}}$. Using the method I described in my post in $\det(AB-BA)=0$? or in $A^2-B^2=\alpha(AB-BA)$

we put $A=uX+vY,B=wX+xY$ where $w=x=2,u=\alpha+i\beta,v=\alpha-i\beta$. Then $AB-BA=(ux-wv)(XY-YX)$, where $ux-wv=4i\beta$. Thus $\det(XY-YX)\not= 0$.

The considered equation becomes $XY=-e^{3i\pi/4}YX$. Consequently $\det(XY-YX)=\det((-e^{3i\pi/4}-1)YX)$ and $\det(YX)=\det(XY)\not= 0$.

Since $\det(XY)=(-1)^ne^{3in\pi/4}\det(XY)$, we deduce that $(-1)^ne^{3in\pi/4}=1$ and $n=0\; mod(8)$.