Find matrice $A_{2 \times 2}$ such $A_{2 \times 2}\in \mathbb{R}$ such that $A^{30}=I$

Question

I have the following question :

Find $A_{2 \times 2}$ matrice $A_{2 \times 2}\in \mathbb{R}$ such that $A^{30}=I$

I tried to solve this problem using determinants using $|A^2|=|A|*|A|$, I think that method is could lead to the answer yet I don't seems to find $A$.

Any ideas?

Any help will be appreciated.

EDIT :

In case $A \neq I$ and the first time $A^k=I$ is for $k=30$

Answer 1

Hint: The rotation matrix $$A = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix}$$ satisfies $$A^k = \begin{bmatrix}\cos k\theta & -\sin k\theta \\ \sin k\theta & \cos k\theta\end{bmatrix}.$$

In other words, applying $k$ rotations by an angle of $\theta$ is the same as one rotation by an angle of $k\theta$.

Also, a rotation matrix is the identity if and only if the angle of rotation is a multiple of $360^{\circ}$.

Using these facts, can you think of an angle $\theta$ such that $A^{30} = I$ but $A^k \neq I$ for $k = 1,2,\ldots,29$?

Answer 2

Hint: Assume that $A = \operatorname{diag}(\lambda_1, \lambda_2)$ is a diagonal matrix. If $A^{30} = I$, what does this tell you about $\lambda_1, \lambda_2$?

Answer 3

There are many solutions, as already pointed out. The simplest one: $A=-I$.

Answer 4

This can be viewed as a group theory problem. We are given that $A$ is a $2\times2$ matrix. Note that $A$ is necessarily invertible because if $A^{30}$ wasn't invertible, then it would simply not equal the identity matrix $I_{2\times2}$ since $det(A^{30})$=$ad-bc$ which is forced to be $1$. We are essentially looking for every $A$ where $ord(A)=30$

Trivially, we have $I, -I$. If we consider diagonal matrices, it suffices to look at the entries of the main diagonal (which are the eigenvalues). Let $\lambda_1$, $\lambda_2$ be the eigenvalues. We want $\lambda_1^{30}=1$ and $\lambda_2^{30}=1$, which are just the roots of unity, but since we are only allowed real entries, we only have $-1,1$.

In the most general setting, we can try to diagonalize $A=P^{-1}DP$ in order to compute $A^{30}$ (or we could use a computer program), but the fact that $A$ is invertible does not guarantee that $A$ is diagonalizable. Even if we consider, invertable, non-diagonalizable matrices, computing $det{(A^{30})}$, which is equivalent to computing the characteristic equation would given us a polynomial of degree $30$, which has no general solution. In other words, it is not possible to find all matrices $A^{30}=I$, although rotation matrices are one non-trivial example, and I am sure there are others.