Rotation matrix and its powers

Question

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?

Answer 1

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.

Answer 2

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 $A^n = A$.

Answer 3

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.