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
 A: We have
$$(A-I)(A^2+A+I)=A^3-I=O\ .$$
Since $A^2+A+I\ne O$, there exists a vector $\bf x$ such that
$${\bf y}=(A^2+A+I){\bf x}$$
is not the zero vector.  Hence $(A-I){\bf y}={\bf0}$, that is,
$$A{\bf y}=\lambda{\bf y}\quad\hbox{with}\quad \lambda=1\ .$$
A: Note that over any field $K$ we have
$A^3 - I = (A - I)(A^2 + A + I), \tag{1}$
and the assumption that $A^2  + A + I \ne 0$ implies the existence of a vector $v$ such that
$w = (A^2 + A + I)v \ne 0; \tag{2}$
then (1) shows that
$(A - I)w = (A - I)(A^2 + A + I)v = (A^3 - I)v = 0, \tag{3}$
since we have $A^3 - I = 0$ by hypothesis.  But (3) yields
$Aw =w, \tag{4}$
that is, $w$ is an eigenvector of $A$ with eigenvalue $1$.  Note in closing that this argument holds over any field $\Bbb K$; the zeroes of $A^2 + A + I$ in $\Bbb K$ or some extension thereof do not enter into the argument.   QED.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: To be explicit, your conclusion that $\lambda=1$ from $\lambda^3=1$ needs to include an argument that $\lambda$ cannot be a complex (non-real) root. That's where the extra hypothesis must be used.
A: 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.  
A: 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.
