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Is there any $2 \times 2$ matrix $A \neq I$ such that $A^3=I$?

In my opinion: no. Thank you very much.

Rodrigo de Azevedo
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Jozef
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2 Answers2

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

GEdgar
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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....

Matthew Towers
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    +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-Álvarez 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. http://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