If $A^3 = A$ then the eigen values are all 1 right?

Question

Since $A^n = PD^nP^{-1}$ where D is a matrix consisting only of the eigenvalues of on its leading diagonal. For the scenario to be true $D^B = D$ which is only true if the eigenvalues are all 1s (not 0s as i think that maybe against the rules, is it?)

Answer 1

I guess the only thing you can say is that the eigenvalues all satisfy the equation $x^3=x$...

Answer 2

take $$A = -I.$$ then $$ A^3 = A. $$ More generally, take a diagonal matrix with all entries $1, -1$ or $0$ and the equation is satisfied. The eigenvalues will have to from that set however.