For what values of $\gamma \in \mathbb{C}$ do there exist non-singular matrices $A , B \in \mathbb{C}^{n \times n}$ such that $$AB = \gamma BA \,?$$

So far what I have done shown that $\gamma$ must be an nth root of unity, by considering the determinant. $$det(AB) = det(\gamma BA)$$ $$det(A)det(B) = \gamma ^n det(B)det(A)$$ Now since both $A$ and $B$ are non singular we have $det(A) \neq 0$ and $det(B) \neq 0$ So:

$$\gamma ^n =1$$. I also know that

$$tr(AB - \gamma BA)=0$$ $$tr(AB) - \gamma tr(BA)=0$$ $$tr(AB)\big(1-\gamma \big) = 0$$ Clearly we can assume $\gamma \neq 1$ since surely we can find $A$ and $B$ such that they commute so we conclude that $tr(AB) = tr(BA) = 0$

Now i'm thinking that we can find matrices $A$ and $B$ for any $\gamma = e^{\frac{2 \pi i}{n}}$ such that $AB = \gamma BA$ but I cannot think of a way of constructing them. Does anyone have any ideas? thanks in advance!

  • $\begingroup$ Examples of the case $\gamma=-1$ are common, for instance $$A=\begin{pmatrix} 0 & 1 \\ 1 & 0\end{pmatrix},\qquad B=\begin{pmatrix} 1&0\\0&-1 \end{pmatrix}.$$ So that leaves complex roots of unity as the interesting case. $\endgroup$ – Semiclassical Aug 14 at 0:49
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    $\begingroup$ Ahah: Equation 7 of this paper gives a construction for arbitrary roots of unity. $\endgroup$ – Semiclassical Aug 14 at 1:00
  • $\begingroup$ wow thank-you! This problem is from a past qualifying exam at my university, I don't think I would have ever come up with such a construction.. $\endgroup$ – Justin Stevenson Aug 14 at 1:07
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    $\begingroup$ Another perspective: $$ AB = \gamma BA \implies A = B(\gamma A)B^{-1}. $$ Consequently, $A$ and $\gamma A$ must have the same eigenvalues. $\endgroup$ – Omnomnomnom Aug 14 at 2:24

For any $n$, define $$B = \text{diag}(e^{2\pi i k / n} : 0 \leq k \leq n-1)$$ and then for any $0 \leq k \leq n-1,$ we can define $A = [a_{i,j}]$ by $$a_{i,j} = \begin{cases}1, & i \equiv j+k\pmod{n} \\ 0, & \text{otherwise}\end{cases}$$

Now just consider the actions of $A$ and $B$ on the standard basis vectors.


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