A $2 \times 2$ matrix $A$ such that $A^n$ is the identity matrix

Question

So basically determine a $2 \times 2$ matrix $A$ such that $A^n$ is an identity matrix, but none of $A^1, A^2,..., A^{n-1}$ are the identity matrix. (Hint: Think geometric mappings)

I don't understand this question at all, can someone help please?

Answer 1

Hint: Rotate through $\frac{2\pi}{n}$.

Answer 2

Any diagonal matrix with entries that are primitive $n$th roots of unity clearly has those properties.

Answer 3

Hint: Consider a rotation by angle $$\frac{2\pi}{n}$$ Apply this rotation $n$ times and you get the identity, but for each $k < n$, applying it $k$ times does not give the identity. Can you write a relevant rotation matrix?

Answer 4

Here's a cute way of looking at it: let

$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$

then

$J^2 = -I. \tag{2}$

From (1) and (2) it follows that

$((\cos \theta)I + (\sin \theta)J)^n = ((\cos n \theta)I + (\sin n \theta)J); \tag{3}$

the proof of (3) is virtually identical to that of de Moivre's formula (see http://en.m.wikipedia.org/wiki/De_Moivre's_formula); indeed, the algebraic maneuvers are essentially the same in either case. Noting that

$(\cos \theta)I + (\sin \theta)J = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end {bmatrix}, \tag{4}$

it follows from (3) that

$\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}^n = \begin{bmatrix} \cos n \theta & -\sin n \theta \\ \sin n \theta & \cos n \theta \end{bmatrix}. \tag{5}$

Now setting $\theta = \frac{2 \pi}{n}$, we let

$A = \begin{bmatrix} \cos \frac{2 \pi}{n} & -\sin \frac{2\pi}{n} \\ \sin \frac{2\pi}{n} & \cos \frac{2\pi}{n} \end{bmatrix}, \tag{6}$

and we see from (5) that

$A^n = \begin{bmatrix} \cos 2\pi & -\sin 2\pi \\ \sin 2\pi & \cos 2\pi \end{bmatrix} = I, \tag{7}$

whereas for $k$, $1 \le k \le n -1$,

$A^k = \begin{bmatrix} \cos \frac{2\pi k}{n} & -\sin \frac{2\pi k}{n} \\ \sin \frac{2\pi k}{n} & \cos \frac{2 \pi k}{n} \end{bmatrix} \ne I, \tag{8}$

establishing the desired result.

Hope this helps. Cheers,

and as always,

Fiat Lux!!!