Show that $\operatorname{rank}(A^2+A+I_3)=1$

If $A \in M_3(\mathbb{R}), A \ne I_3$ and $A^3=I_3$ Show that $\operatorname{rank}(A^2+A+I_3)=1$.

What I have reached so far is that $\operatorname{rank}(A-I_3)+\operatorname{rank}(A^2+A+I_3)\le 3$ using Sylvester theorem, and also it can be quite easily proven that $\operatorname{rank}(A^2+A+I_3) \ne 0$, if that helps in any way, but I have no idea what should I do now.
I've seen this problem statement in the archives of a contest, but no solving is provided.

Can anyone help me with this?

P.S. It is not an actual homework, rather a problem to train for future contests, but I think it can be somehow related to the tag

-

Since you know that $A^3=I$, we have $m_A(x)|x^3-1=(x-1)(x^2+x+1)$ (the minimal polynomial). Since the caracteristic polynomial and the minimal have the same irriducible factors, $m_A(x)\neq x^2+x+1$, since otherwise $p_A(x)=(x^2+x+1)^k$, but $\deg(p_A(x))=3$.
Since $A\neq I$, $m_A(x)\neq (x-1)$. Hence all your information is equivalent to $m_A(x)=x^3-1=(x-1)(x^2+x+1)$. Hence $A$ is diagonalizable over $\mathbb{C}$ to the form $D=diag(1,\alpha,\beta)$ where $\alpha,\beta$ are the roots of $x^2+x+1=0$.
Write $A=PDP^{-1}$. Then $A^2+A+I=P(D^2+D+I)P^{-1})$ and $rank(D^2+D+I)=1$ (you should see easily why). Hence $rank(A^2+A+I)=1$.

-
How do I know if a matrix is diagonalizable or not? – Bujanca Mihai Mar 5 '13 at 17:17
For example, if all the eigenvalues of $A$ are distinct then $A$ is diagonalizable. Read more here: en.wikipedia.org/wiki/Diagonalizable_matrix – Dennis Gulko Mar 5 '13 at 17:25

$A$ is a root of $x^{3}-1$, which has distinct roots, so $A$ can diagonalized over the complex numbers. Since it is a real matrix, and it is different from the identity, the diagonal form must be $$B =\begin{bmatrix} 1&&\\ &\omega&\\ &&\omega^{2} \end{bmatrix},$$ where $\omega$ is a primitive third root of unity. Since $1 + \omega + \omega^{2} = 0$, we have that $$B^{2} + B + I = \begin{bmatrix} 3&&\\ &0&\\ &&0 \end{bmatrix}$$ has rank one. Now $A$ and $B$ are conjugate, and conjugacy preserves rank.

-
How do I know if a matrix is diagonalizable or not? – Bujanca Mihai Mar 5 '13 at 17:17
@BujancaMihai, I have used the criterion that if the minimal polynomial has all roots in the field, and they are distinct, then the matrix is diagonalizable. – Andreas Caranti Mar 5 '13 at 18:39

Hint: What can you say about the minimal polynomial of $A$?

-
The text passage contains a mathematical problem statement and a detailed solution approach, including a question about diagonalizability of matrices.
single_question_with_answer
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
Show that $\operatorname{rank}(A^2+A+I_3)=1$

If $A \in M_3(\mathbb{R}), A \ne I_3$ and $A^3=I_3$ Show that $\operatorname{rank}(A^2+A+I_3)=1$.

What I have reached so far is that $\operatorname{rank}(A-I_3)+\operatorname{rank}(A^2+A+I_3)\le 3$ using Sylvester theorem, and also it can be quite easily proven that $\operatorname{rank}(A^2+A+I_3) \ne 0$, if that helps in any way, but I have no idea what should I do now.
I've seen this problem statement in the archives of a contest, but no solving is provided.

Can anyone help me with this?

P.S. It is not an actual homework, rather a problem to train for future contests, but I think it can be somehow related to the tag

-

Since you know that $A^3=I$, we have $m_A(x)|x^3-1=(x-1)(x^2+x+1)$ (the minimal polynomial). Since the caracteristic polynomial and the minimal have the same irriducible factors, $m_A(x)\neq x^2+x+1$, since otherwise $p_A(x)=(x^2+x+1)^k$, but $\deg(p_A(x))=3$.
Since $A\neq I$, $m_A(x)\neq (x-1)$. Hence all your information is equivalent to $m_A(x)=x^3-1=(x-1)(x^2+x+1)$. Hence $A$ is diagonalizable over $\mathbb{C}$ to the form $D=diag(1,\alpha,\beta)$ where $\alpha,\beta$ are the roots of $x^2+x+1=0$.
Write $A=PDP^{-1}$. Then $A^2+A+I=P(D^2+D+I)P^{-1})$ and $rank(D^2+D+I)=1$ (you should see easily why). Hence $rank(A^2+A+I)=1$.

-
How do I know if a matrix is diagonalizable or not? – Bujanca Mihai Mar 5 '13 at 17:17
For example, if all the eigenvalues of $A$ are distinct then $A$ is diagonalizable. Read more here: en.wikipedia.org/wiki/Diagonalizable_matrix – Dennis Gulko Mar 5 '13 at 17:25

$A$ is a root of $x^{3}-1$, which has distinct roots, so $A$ can diagonalized over the complex numbers. Since it is a real matrix, and it is different from the identity, the diagonal form must be $$B =\begin{bmatrix} 1&&\\ &\omega&\\ &&\omega^{2} \end{bmatrix},$$ where $\omega$ is a primitive third root of unity. Since $1 + \omega + \omega^{2} = 0$, we have that $$B^{2} + B + I = \begin{bmatrix} 3&&\\ &0&\\ &&0 \end{bmatrix}$$ has rank one. Now $A$ and $B$ are conjugate, and conjugacy preserves rank.

-
How do I know if a matrix is diagonalizable or not? – Bujanca Mihai Mar 5 '13 at 17:17
@BujancaMihai, I have used the criterion that if the minimal polynomial has all roots in the field, and they are distinct, then the matrix is diagonalizable. – Andreas Caranti Mar 5 '13 at 18:39

Hint: What can you say about the minimal polynomial of $A$?

-
user
The text passage contains a mathematical problem statement and a detailed solution approach, including a question about diagonalizability of matrices.

TYPE: SINGLE_QUESTION_WITH_ANSWER