The vector $[4 - 2 \sqrt{3}, 4 + 8 \sqrt{3}]^T$ is the image of a vector $v$ under the linear transformation $F : R^2 \rightarrow R^2,$ which is a linear transformation which first rotates a vector anti-clockwise by $\pi/3$ and then halves the first element of the vector $v.$ Calculate $v.$
What I have tried, as we know the operator that rotates a vector in $R^2$ through a given angle $\theta$ is called a rotation operator in $R^2$ and here $F(v) = (v_1 cos \theta - v_2 sin \theta; v_1 sin \theta + v_2 cos \theta)$ and $\theta = \pi/3.$ So $F(v)$ becomes $(\frac{v_1}{2} - \frac{\sqrt3 v_2}{2}; \frac{\sqrt3v_1}{2} + \frac{v_2}{2}).$ Then according to the question halves the first element of the vector $v.$ Hence
$F(v) = (\frac{v_1}{4} - \frac{\sqrt3 v_2}{2}; \frac{\sqrt3v_1}{4} + \frac{v_2}{2}) = [4 - 2 \sqrt{3}, 4 + 8 \sqrt{3}].$ Is it correct? Kindly let me know.