If A is an n x n matrix over R such that A^3 = A + 1, prove that det(A) > 0 .
The Attempt at a Solution
So, what I've done is factor the expression to get A(A+1)(A-1) = 1, then taking the determinant of both sides, I get det(A)det(A+1)det(A-1) = 1. I thought that maybe I could argue that maybe if det(A) < 0, then this would lead to a contradiction, but that doesn't seem to be going anywhere.
Should I try to solve it without the explicit use of determinants?