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
Is there any matrix $2\times 2$ such that $A\neq I$ but $ A^3=I$
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linear-algebra
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0See also: [How to construct a $2\times 2$ real matrix $A$ not equal to Identity such that $A^3=I$?](https://math.stackexchange.com/q/182632) – 2017-10-03
2 Answers
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Rotation by $2\pi/3$ in the plane. Find the $2 \times 2$ matrix that gives you this linear transformation.
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1Ok, I'll think. thanks – 2011-08-20
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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....
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0Yes, $2\times 2$ 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). – 2011-08-20