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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 $A^{17} = I_2$.

Squirrel-Power
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    Hint: consider the representation of your matrices as acting on $\mathbb{R}^2$. Can you think of an operation such that performing it 17 times is the same as the identity? How about a suitable rotation? – Steven Stadnicki Jan 27 '17 at 00:15
  • What about $$A=\begin{pmatrix} 1 & 1 \ 0 & 1 \end{pmatrix}$$? – Jack D'Aurizio Jan 27 '17 at 00:57

2 Answers2

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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.

Arthur
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  • Thanks. but why can we represent a complex number as you described? – Squirrel-Power Jan 27 '17 at 00:19
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    @User90024 Clearly, under addition, the set of such matrices is a $2$-dimensional vector space over $\Bbb R$. The only thing missing is whether the representation gives the correct multiplication, which is easily checked by hand. – Arthur Jan 27 '17 at 00:23
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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 $A^{17}=I$.

carmichael561
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