Rotation matrix and its powers A  is a rotation matrix in  ${\mathbb{R}}^2$ by angle $θ=30°$.
I know the matrix:
$A= \begin{bmatrix} \frac{\sqrt{3}}{2} & -\frac{1}{2}\\\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$
I can write $A$ to another form:
$$
A=
\begin{bmatrix}
\frac{\sqrt{3}}{2} \\
-\frac{1}{2} \\
\frac{1}{2} \\
\frac{\sqrt{3}}{2}\\
\end{bmatrix}
$$
The question is how many of $A$'s powers are linearly independent?
So i have $A$ , $A^2$, $A^3$, ... , $A^{12} =$ identity matrix
$A = -A^{7}$
$A^2 = -A^{8}$
.
.
.
$A^6 = - A^{12}$
So I know that the numbers of linearly independent vectors $\leqslant 6$.
As these are 4 dimensional vectors, there are maximum 4 linearly dependent ???
How many of $A$'s powers are linearly independent?
 A: The characteristic equation for matrix $A$ is $t^2-\sqrt 3t+1=0$.  Using Cayley-Hamilton theorem
$A^2-\sqrt 3 A+I_3=0$
$\implies A^2=\sqrt 3 A-I_3$.....(*)
So $A^2$ is linear combination of $A$ and $A^0$. Multiply (*) by $A$ repeatedly to observe that each $A^n, (n\ge 2)$ is linear combination of $A$ and $A^0$. So, only $A$ and $A^0$ are LI.
A: A rotation matrix in two dimensions has the form
$$
\pmatrix{\cos\phi&\sin\phi\\-\sin\phi&\cos\phi}\;.
$$
Note that the second column is a linear function of the first column: The second column vector is the first column vector rotated by $\frac\pi2$. Thus the entire rotation matrix is a linear function of its first column. It follows that no more than $2$ rotation matrices can be linearly independent. Clearly $A^0$ and $A^1$ are linearly independent; thus your set contains two linearly independent matrices.
A: The rotation matrix in 2D has the following property:
$$
\pmatrix{\cos\phi&\sin\phi\\-\sin\phi&\cos\phi}^n = \pmatrix{\cos n\phi&\sin n\phi\\-\sin n\phi&\cos n\phi}
$$
You can prove this using mathematical induction. Using this formula, you can find the smallest positive integer $n$ that makes $A^n = A$.
