Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

3 Answers 3

up vote 3 down vote accepted

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 $rank(A^2+A+I)=1$.

share|improve this answer
    
How do I know if a matrix is diagonalizable or not? –  Bujanca Mihai Mar 5 '13 at 17:17
1  
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.

share|improve this answer
    
How do I know if a matrix is diagonalizable or not? –  Bujanca Mihai Mar 5 '13 at 17:17
1  
@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 $A$?

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.