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.

I want to ask you this question: Is there any matrix $2\times 2$ such that $A\neq I$ but $A^3=I$. In my opinion: No. Thank you very much

share|improve this question
21  
I don't think it is a matter of opinions... –  Álvaro Lozano-Robledo Aug 20 '11 at 15:42
add comment

2 Answers

up vote 25 down vote accepted

Rotation by $2\pi/3$ in the plane. Find the $2 \times 2$ matrix that gives you this linear transformation.

share|improve this answer
    
Hi! How did you get that? –  Jozef Aug 20 '11 at 15:08
12  
Imagine something when applied 3 times is the identity. How do you think? –  GEdgar Aug 20 '11 at 15:12
1  
Ok, I'll think. thanks –  Jozef Aug 20 '11 at 15:20
add comment

Since you don't specify what field the entries of this matrix have to come from, I could just take a diagonal matrix whose entries are $1$ and $\omega$ where $\omega=e^{2\pi i /3}$ is a primitive cube root of 1 in the complex numbers.

I guess you want real or integer entries though. If $A^3=I$ then the eigenvalues of $A$, that is, the roots of the characteristic polynomial, have to be third roots of unity. A primitive third root of unity satisfies $x^2+x+1=0$, so you could look for a matrix over the integers with that as a characteristic polynomial....

share|improve this answer
2  
+1. Nice! Your argument shows that this holds for any (nonzero) commutative ring. –  Pierre-Yves Gaillard Aug 20 '11 at 16:17
    
Another way to get an integral matrix that does the job it to take the automorphism of order $3$ of the torus, then the induced map on homology. In fact, there's an automorphism of order $6$ coming from the symmetry of the hexagonal tiling. –  Ryan Budney Aug 20 '11 at 16:35
    
@Ryan: over $\mathbb Z$, you have reduced to the problem to the (equivalent, I'm pretty sure---given enough technology) one of finding an automorphism of order $3$ of the torus :) –  Mariano Suárez-Alvarez Aug 20 '11 at 22:07
    
Yes, $2\times 2$ matrices of the integers of finite order all come from symmetries of the torus. This is a special case of what's called the Nielsen Realization problem -- which is solved, by-the-way. en.wikipedia.org/wiki/Nielsen_realization_problem Among other things, this gives you a fairly intuitive way to enumerate all the finite-order elements of $GL_2 \mathbb Z$ (up to conjugacy). –  Ryan Budney Aug 20 '11 at 22:20
add comment

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.