There is a natural $n,$ such that $n≥2$ where $A, B$ are square matrices of order n, idempotents, and that det($A$$-$$B$)$≠$$0$
I tried to use the idempotence propriety, so I thought about starting from here,continuing with this and proving that it's not equal to zero.
I know that (A+B)(A+B-I)=AB+BA and (A-B)(A+B-I)=AB-BA but I don't know how to actually proof that their ranks are equal.
I would be more than happy to receive some ideas.
Hint: $$ (2I-A-B)(AB-BA) = AB-BA+ABA-BAB = (A-B)(AB+BA) \\ (2I-A-B)(AB+BA) = AB+BA-ABA-BAB = (A-B)(AB-BA) $$ Now you can show that each element of the kernel of (the linear map represented by) $AB+BA$ is also an element of the kernel of (the linear map represented by) $AB-BA$ and vice versa, which in turn shows the equality of the ranks.
Note that you can use $(A-B)^{-1}$ but not $(2I-A-B)^{-1}.$