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

Since $A^3=I$, therefore $p(x)=x^3-1$ is an annihilating polynomial for the matrix $A$. Let $m(x)$ be the minimal polynomial for matrix $A$, then $m(x)$ must divide $p(x)$. Since $p(x)=(x-1)(x^2+x+1)$, this means either $m(x)=p(x)$ or $m(x)=x-1$ or $m(x)=x^2+x+1$. From the hypothesis given we know that $m(x) \neq x^2+x+1$. Thus $m(x)$ will satisfy either of the remaining possibilities. In both cases we get that $\lambda=1$ is an eigen value.
