Question :
Let $R$ be the rotation around the origin clockwise by $\frac{ \pi}{3}$. Is $R$ a linear transformation from $\Bbb R^2 $ to $\Bbb R^2$? Write down the corresponding matrix $Au = R(u)$. Compute $R(u)$ with $u=[1,3]$
My approach:
$$R(u)=\underbrace{\begin{pmatrix}\cos \theta&\sin \theta\\-\sin \theta &\cos \theta\end{pmatrix}}_{\text { rotation matrix for clockwise direction}}\vec u$$
$$ R
\begin{bmatrix}
1 \\ 3
\end{bmatrix} = \begin{pmatrix}\cos \frac{ \pi}{3} &\sin \frac{ \pi}{3}\\-\sin \frac{ \pi}{3} &\cos \frac{ \pi}{3}\end{pmatrix} \begin{bmatrix}
1 = \vec u_{1} \\ 3 = \vec u_{2}
\end{bmatrix} $$
Product of matrices :
$$ R \begin{bmatrix}
1 \\ 3
\end{bmatrix} = \begin{pmatrix}\cos \frac{ \pi}{3} \cdot 1 ~~~+&\sin \frac{ \pi}{3} \cdot 3\\-\sin \frac{ \pi}{3} \cdot 1 ~~~+&\cos \frac{ \pi}{3} \cdot 3\end{pmatrix}$$
$$ \underbrace{R
\begin{bmatrix}
1 \\ 3
\end{bmatrix}}_{\text{R(u)}}= \underbrace{\begin{pmatrix} \frac{1+3 \sqrt{3}}{2}\\ \frac{3-\sqrt{3}}{2}\end{pmatrix}}_{\text{A x u}} $$
Conclusion :
$R $ is a linear transformation.
I feel I understand the Question, but I was hesitant on following wording
— “is R a linear transformation ?"
Could guys please validate the my answer and elaborate a little on the Wording.
Edit :
$ \vec u = \begin{bmatrix}
x_1\\ x_2
\end{bmatrix} $ , then a clockwise rotation by $\frac{\pi}{2}$ gives $\vec w = \begin{bmatrix}
x_2\\ -x_1
\end{bmatrix}$
Using trigonometry:
$$ R(\vec u) = \cos \left(\frac{\pi}{2}\right) ~\vec u + \sin \left(\frac{\pi}{2}\right)~\vec w$$
Thus, a general form would be :
$$ R(\vec u) = (\cos \theta) ~\vec u + (\sin \theta)~\vec w$$
Using $\theta = \frac{\pi}{3}$ in clockwise direction :
$$ = \cos \left(\frac{\pi}{3}\right) \begin{bmatrix}
x_1 \\ x_2
\end{bmatrix} + \sin \left(\frac{\pi}{3}\right) \begin{bmatrix}
x_2 \\ -x_1
\end{bmatrix} $$
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) \cdot x_1 + \sin \left(\frac{\pi}{3}\right) \cdot x_2\\ \cos \left(\frac{\pi}{3}\right) \cdot x_2 – \sin \left(\frac{\pi}{3}\right) \cdot x_1
\end{bmatrix} $$
Rearranging :
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) \cdot x_1 + \sin \left(\frac{\pi}{3}\right) \cdot x_2\\ – \sin \left(\frac{\pi}{3}\right) \cdot x_1 +
\cos \left(\frac{\pi}{3}\right) \cdot x_2 \end{bmatrix} $$
$$ = \begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) & \sin \left(\frac{\pi}{3}\right) \\ – \sin \left(\frac{\pi}{3}\right) & \cos \left(\frac{\pi}{3}\right)
\end{bmatrix} ~ \begin{bmatrix}
x_1\\ x_2
\end{bmatrix} $$
$$ R(\vec u) = \underbrace{\begin{bmatrix}
\cos \left(\frac{\pi}{3}\right) & \sin \left(\frac{\pi}{3}\right) \\ – \sin \left(\frac{\pi}{3}\right) & \cos \left(\frac{\pi}{3}\right)
\end{bmatrix}}_{\text { rotation matrix}}~ \vec u$$
Hence proved it’s linear because it gives the form
$$ R(\vec u) = A \cdot ~\vec u$$
Lastly :
$$ \underbrace{R
\begin{bmatrix}
1 \\ 3
\end{bmatrix}}_{\text{R(u)}}= \underbrace{\begin{pmatrix} \frac{1+3 \sqrt{3}}{2}\\ \frac{3-\sqrt{3}}{2}\end{pmatrix}}_{\text{A x u}} $$
Best Answer
Your approach is correct for the last two questions. However, you have not proven that $R$ is a linear transformation. I will answer your question with detail.
A linear transformation from a $K$-vector space $V$ to another $K$-vector space $W$ is a function $R:V\to W$ such that the following two properties hold:
In your case, in order to prove that $R\colon \mathbb{R}^2\to\mathbb{R}^2$ is a linear transformation, you just have to prove that the properties 1. and 2. hold with $V,W=\mathbb{R}^2$, $K=\mathbb{R}$ and arbitrary vectors $v,w$ (as @Salihcyilmaz has pointed out in a comment). If you have already proven that $R(v)=Av$, then said properties are immediate to prove, since:
P.S.: Note that we have only made use of the fact that there exists a fixed matrix $A$ such that $R(v)=Av$ for all $v\in\mathbb{R}^2$. Therefore, any transformation which can be represented by a matrix (in this sense) is a linear transformation.