Give $2$ matrices $A$ and $B$, that are real, with size $n$($n\ge 2$) and satisfy $AB- nA - B$ = $0$. Show that $AB^{2n} = B^{2n}A$

Question

Give $2$ matrices $A$ and $B$ that are real with size $n$ ($n\ge 2$) and satisfy $AB- nA - B$ = $0$. Show that $AB^{2n}$ = $B^{2n}A$ (Both $A$ and $B$ are nonsingular matrices)

(this comes from a mocking test version of the final exam of my linear algebra course, and normally the hardest task to categorize the level of the learner)

I have tried to use algebraical manipulation, but it doesn't work, and I tried to use mathematical induction, but it didn't work either. Can anyone help me with this question?

Thanks in advance

Answer 1

Notice that $$nI = AB-nA - B +nI = A(B-nI) - (B-nI) = (A-I)(B-nI)$$ so by Binet-Cauchy theorem we also have $$nI = (B-nI)(A-I) = BA-nA-B+nI \implies BA-nA-B = 0.$$ But since $AB-nA - B=0$, it has to be $AB = BA$. Therefore in particular $AB^{2n} = B^{2n}A$ holds.