Answer
a) The matrix corresponding to linear transformation $T$ is given by:
$\left[ {\begin{array}{*{20}{c}} { \frac{1}{{2}}}&{\frac{\sqrt{3}}{2}}\\ -{\frac{\sqrt{3}}{2}}&{\frac{1}{2}} \end{array}} \right]$
b) Hence, 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 the 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 $\theta =60^{\circ}$, we have
$T(x,y)=(x\cos( 60^{\circ})+y\sin(60^{\circ}) ,{ }-x\sin(60^{\circ}) +y\cos(60^{\circ})) \\ =[x(\frac{1}{2})+y(\frac{\sqrt{3}}{2}) ,{ }-x(\frac{\sqrt{3}}{2}) +y(\frac{1}{2})] $
Take the standard basis for $R^2,$ which is $\{(1,0),(0,1)\}=\{e_1,e_2\}$.
Then, $T(1,0)=[1(\frac{1}{2})+0(\frac{\sqrt{3}}{2}) ,{ }-1(\frac{\sqrt{3}}{2}) +0(\frac{1}{2})]=(\frac{1}{{2}},-\frac{\sqrt{3}}{2})\\ T(0,1)=[0(\frac{1}{2})+1(\frac{\sqrt{3}}{2}) ,{ }-0(\frac{\sqrt{3}}{2}) +1(\frac{1}{2})]=(\frac{\sqrt{3}}{2},\frac{1}{{2}}) $
Therefore, the matrix corresponding to linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} { \frac{1}{{2}}}&{\frac{\sqrt{3}}{2}}\\ -{\frac{\sqrt{3}}{2}}&{\frac{1}{2}} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 1\\ {2} \end{array}} \right]$ is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 1\\ {2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { \frac{1}{{2}}}&{\frac{\sqrt{3}}{2}}\\ -{\frac{\sqrt{3}}{2}}&{\frac{1}{2}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ {2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { \frac{1}{2}+\sqrt{3}}\\ { -\frac{\sqrt{3}}{2}+1} \end{array}} \right] \end{array}$
c) See image.
