Does the fact that $A^{17} = I_2$ imply that the matrix $A$ must be $I_2$?
Since the question does not specify whether the entries of $A$ are allowed to be complex valued functions, I said that IF the entries can indeed be complex numbers, then the answer must be no, because we can take $A = (\cos(\frac{2\pi k}{17})+i \sin(\frac{2\pi k}{17}))I_2, k = 0,1,...,16$. (i.e. the identity matrix times the complex root of the equation $x^{17}=1$)
I was wondering whether, if the entries are only taken over real numbers, there exist $A$ that satisfy the condition $A^{17} = I_2$.
$\endgroup$ 22 Answers
$\begingroup$You can actually represent complex numbers as $2\times2$ real matrices! The complex number $a+bi$ is represented as $\left(\begin{smallmatrix}a&-b\\b&a\end{smallmatrix}\right)$. So just take any $17$th root of unity, insert its real and imaginary parts into that matrix, and you have a matrix which is a seventeenth root of $I_2$. This is the same answer as the other one, but with a slightly different interpretation.
$\endgroup$ 2 $\begingroup$Yes, define $$ A=\begin{bmatrix}\cos\Big(\frac{2\pi}{17}\Big)&-\sin\Big(\frac{2\pi}{17}\Big)\\\sin\Big(\frac{2\pi}{17}\Big)&\cos\Big(\frac{2\pi}{17}\Big)\end{bmatrix}$$ $A$ is a counterclockwise rotation by $\frac{2\pi}{17}$ radians, hence $A^{17}=I$.
$\endgroup$
