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Let $A, B \in M_n(\mathbb{R})$ , $\alpha\in\mathbb{R}$, such that $A^2-B^2=\alpha(AB-BA)$. Prove that

$a)$ If $\alpha=0$ and $n$ odd, then $\det(AB-BA)=0$

$b)$ If $\alpha\neq0$ then $(AB-BA)^n=0_n$

For $a)$ we use the fact that $$\det(A+B)(A-B)=\det(A-B)(A+B)$$ which means that $$\det(AB-BA)=\det(-(AB-BA))$$ and since $n$ is odd we obtain the conclusion. The second point is, however, a little bit trickier. I managed to show just that $\det(AB-BA)=0$. Using the same method as for $a)$, we observe that $$\det((\alpha+1)(AB-BA))=\det((\alpha-1)(AB-BA))$$ and since $\alpha\neq0$, we obtain that our determinant is $0$, but from here I don't have any idea what should I do next.

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razvanelda
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If $\alpha\not= 0$, then we show that $A,B$ are simultaneously triangularizable (denoted by ST) over $\mathbb{C}$; that implies that $AB-BA$ is nilpotent.

Let $A=uX+vY,B=wX+xY$; if $ux-wv\not= 0$, then it suffices to show that $X,Y$ are ST. We obtain

$(u^2-w^2)X^2+(v^2-x^2)Y^2+(?)XY+(?)YX=0$.

Putting $u=w=x=1,v=-1$, we obtain $(1+\alpha)XY=(\alpha -1)YX$.

Case 1. $\alpha=\pm 1$. Then $XY=0$ or $YX=0$ and $X,Y$ are ST.

Case 2. $\alpha\notin \{0,\pm 1\}$. Then $XY=\dfrac{\alpha -1}{\alpha +1}YX=kYX$.

Since $\alpha\in\mathbb{R}$, $k$ cannot be a primitive root of unity; according to a result from Drazin, $X,Y$ are ST and we are done.

cf. my post in

$\det(AB-BA)=0$?

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    Just a remark: no need to use that $X,Y$ are ST. After the substitution $A=X-Y$ and $B=X+Y$ we get $AB-BA=2\alpha(XY-YX)$. In Case 1. notice that $XY=0$ implies $(YX)^2=0$, hence $YX$ is nilpotent, and vice versa. In Case 2. we have $AB-BA=2\alpha(k-1)YX$, and since $XY=kYX$ and $XY,YX$ have the same spectra we get that $XY,YX$ are nilpotent. – user26857 Jan 09 '20 at 16:30
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    @user26857 , you are right, but "$A,B$ are ST" is much stronger than the asked question. Indeed, if $A,B$ are triangular and the diagonal of $A$ is $(\lambda_i)_i$, then the diagonal of $B$ is $(\pm \lambda_i)_i$. –  Jan 09 '20 at 21:44
  • I can agree with this. – user26857 Jan 09 '20 at 22:20