How to solve matrix polynomial, that is $A^{15}$ $A$ is a matrix that suffice $A^2 = A - I$ ($I$ is identity matrix.)
Of course, $A^{15} = {(A^2)}^7A$
I think i have to find the elements of matrix A. I tried using completing the square,  $$A^2 - A + I = 0$$
$${(A-I/2)}^2 + (3/4) I = 0$$
Which leads to imaginary number.

$A$ is a zero of $x^2 -(x - 1)=x^2-x+1$, which you may recognize as the $6$-th cyclotomic polynomial. Therefore, $A^6 = I$ and so $A^{15} = A^{12} A^3 = A^3$. Now, $A^3 = A^2-A = -I$.
A systematic way that does not require bright ideas is to use polynomial division. The quotient is irrelevant; only the remainder matters:
$$
x^{15} = (x^{13} + x^{12} - x^{10} - x^9 + x^7 + x^6 - x^4 - x^3 + x + 1) (x^2 -(x - 1))-1
$$
Therefore, $A^{15} = -I$.
