Answer
a) The matrix corresponding to the linear transformation $T$ is given by:
$\left[ {\begin{array}{*{20}{c}} {\frac{\sqrt{3}}{2}}&{ \frac{1}{{2}}}&\\ -{\frac{1}{2}}&{\frac{\sqrt{3}}{2}} \end{array}} \right]$
b) The image of $v=\left[ {\begin{array}{*{20}{c}} 1\\ {2} \end{array}} \right]$ is $ \left[ {\begin{array}{*{20}{c}} { \frac{1}{2}+\sqrt{3}}\\ { \frac{\sqrt{3}}{2}-1} \end{array}} \right] $
c) See image.
Work Step by Step
The linear transformation corresponding to counterclockwise rotation at an angle $\theta$ is given by
$T(x,y)=(x\cos\theta +y\sin\theta ,{ }-x\sin\theta +y\cos\theta) $
a) When $\theta =30^{\circ}$ $T(x,y)=(x\cos( 30^{\circ})+y\sin(30^{\circ}) ,{ }-x\sin(30^{\circ}) +y\cos(30^{\circ})) \\ =[x(\frac{\sqrt{3}}{2})+y(\frac{1}{2}) ,{ }-x(\frac{1}{2}) +y(\frac{\sqrt{3}}{2})] $
Take the standard basis for $R^2,$ which is $\{(1,0),(0,1)\}=\{e_1,e_2\}$.
Then, $T(1,0)=[1(\frac{\sqrt{3}}{2})+0(\frac{1}{2}) ,{ }-1(\frac{1}{2}) +0(\frac{\sqrt{3}}{2})] =(\frac{\sqrt{3}}{2} ,{ }-\frac{1}{2}) \\ T(0,1)=[0(\frac{\sqrt{3}}{2})+1(\frac{1}{2}) ,{ }-0(\frac{1}{2}) +1(\frac{\sqrt{3}}{2})] =(\frac{1}{2} ,{ }\frac{\sqrt{3}}{2}) $
Therefore, the matrix corresponding to linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} {\frac{\sqrt{3}}{2}}&{ \frac{1}{{2}}}&\\ -{\frac{1}{2}}&{\frac{\sqrt{3}}{2}} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 1\\ {2} \end{array}} \right]$ is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 2\\ {1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {\frac{\sqrt{3}}{2}}&{ \frac{1}{{2}}}&\\ -{\frac{1}{2}}&{\frac{\sqrt{3}}{2}}\end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ {1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { \frac{1}{2}+\sqrt{3}}\\ { \frac{\sqrt{3}}{2}-1} \end{array}} \right] \end{array}$
c) See image.
