Let $A\in M_{n\times n}(\mathbb{R})$ with $A^2+A+I_n =0_n$. Prove that $A^3=I_n$

Question

Let $A\in M_{n\times n}(\mathbb{R})$ with $A^2+A+I_n =0_n$. Prove that $A^3=I_n$.

Progress:
I managed to prove that $A$ is invertible. Checked out this, with no help yet.

Answer 1

Hint:

remember:

$$ A^3-I=(A-I)(A^2+A+I) $$

Answer 2

Times by $A$ to get $$A^3 + A^2 + A = 0$$ so $$A^3 = -A^2 - A$$ But we know from the given condition that $I_n = - A^2 - A$, so $A^3 = I_n$.

This is quite a typical trick used in the application of what is known as the Cayley-Hamilton Theorem.

Answer 3

$$A^2+A+I=0\Rightarrow A^2+A=-I$$ Therefore $$A(A^2+A+I)=A\cdot 0=0\Rightarrow A^3+A^2+A=0\Rightarrow A^3+(-I)=0\Rightarrow A^3=I$$