Show that $1$ is an eigenvalue of $A$ Let $A \in \mathbb{K}_{n}$. Suppose that $A^3=I_{n}$ and $I_{n}+A+A^2\neq O_{n}$. Where $O_{n}$ is the null matrix. Show that $1$ is an eigenvalue of $A$.
I couldn't show what is ask by using all the hypothesis. My work was:
Let $v \in \mathbb{K}_{nx1}$, and $\lambda \in \mathbb{K}$. So one have $Av=\lambda v$ and also,
$$A^2v=\lambda A v$$ 
$$A^3v=\lambda A^2 v$$
By hypothesis, $A^3=I_{n}$. So,
$$I_{n}v=\lambda A^2v$$
$$v=\lambda A^2v$$
$$v=\lambda^2 Av$$
$$v=\lambda^3 v$$
So, $\lambda=1$. My doubt is where $I_{n}+A+A^2\neq O_{n}$ hypothesis is need. Thanks

Assume by contradiction that 1 is not an eigenvalue of $A$. Then $\det(A-I) \neq 0$, and therefore $A-I$ is invertible.
Hence
$$A^2+A+I=(A^3-I)(A-I)^{-1}=0$$
contradiction.  
