How to solve matrix polynomial, that is $A^{15}$

Question

$A$ 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.

Answer 1

From $A^2=A-I$ we have $$A^3=AA^2=A(A-I)=A^2-A=A-I-A=-I$$ thus $$A^{15}=(A^3)^5=(-I)^5=-I$$

Answer 2

$A$ 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, $A^{15} = -I$.