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I found the question in an online a source of challenging linear algebra problems, unfortunately there are no answers.

Question: Find a real $2\times 2$ matrix $A$ (other than $A = I$ ) such that $A^5 = I$.

I'm beginning to think no such matrix exists, but the way the question is posed it doesn't seem they would pull a trick like that.

user127.0.0.1
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Grid
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2 Answers2

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Here is a rotation matrix $$\begin{pmatrix}\cos t &-\sin t\\\sin t&\cos t\end{pmatrix}$$ that does counterclockwise rotation on the usual $xy$-plane by an angle of $t$. Now, put $t=2\pi/5$ to get what you want (Rotating $5$ times by an angle of $2\pi/5$ should mean no rotation (and thus an identity matrix)).

Lord Soth
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We can identify a complex number $a+ib$ by the matrix $$\begin{pmatrix}a&-b\\b&a\end{pmatrix}$$ Can you find $z\in\Bbb C$ such that $z^5=1$?

Pedro
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  • The question asks about a real matrix though. – Grid Jan 26 '14 at 23:39
  • @Grid $a$ and $b$ are (implicitly) real. – Git Gud Jan 26 '14 at 23:39
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    The matrix will be real, @Grid since we're using $\Re z$ and $\Im z$. – Pedro Jan 26 '14 at 23:39
  • One z would be (-1)^(2/5) which is approximately 0.309 + 0.951i so the a would be around .309 and cos(72°) is approximately .309, so this seems intimately related to the rotations. – Grid Jan 26 '14 at 23:51
  • @Grid Of course. $e^{i\theta}$ rotates complex numbers. Do you know that $\omega_k=e^{2\pi i k/5}$ for $k=0,1,2,3,4$ is such that $\omega_k^5=1$? – Pedro Jan 26 '14 at 23:56
  • @PedroTamaroff Thanks. Vaguely knew the connections, but makes much more sense when explicitly written out. – Grid Jan 27 '14 at 00:09