Answer
a) The matrix corresponding to linear transformation $T$ is given by:
$\left[ {\begin{array}{*{20}{c}} { \frac{1}{\sqrt{2}}}&{-\frac{1}{\sqrt{2}}}\\ \frac{1}{\sqrt{2}}&{\frac{1}{\sqrt{2}}} \end{array}} \right]$
b) The image of $v=\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right]$ is $\left[ {\begin{array}{*{20}{c}} { 0}\\ { 2\sqrt{2}} \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 $\theta =45^{\circ}$, we have
$T(x,y)=(x\cos( 45^{\circ})-y\sin(45^{\circ}) ,{ }x\sin(45^{\circ}) +y\cos(45^{\circ})) \\ =[x(\frac{1}{\sqrt{2}})-y(\frac{1}{\sqrt{2}}) ,{ }x(\frac{1}{\sqrt{2}}) +y(\frac{1}{\sqrt{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}{\sqrt{2}})-0(\frac{1}{\sqrt{2}}) ,{ }1(\frac{1}{\sqrt{2}}) +0(\frac{1}{\sqrt{2}})]=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})\\ T(0,1)=[0(\frac{1}{\sqrt{2}})-1(\frac{1}{\sqrt{2}}) ,{ }0(\frac{1}{\sqrt{2}}) +1(\frac{1}{\sqrt{2}})]=(-\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}) $
Therefore, the matrix corresponding to linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} { \frac{1}{\sqrt{2}}}&{-\frac{1}{\sqrt{2}}}\\ \frac{1}{\sqrt{2}}&{\frac{1}{\sqrt{2}}} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right]$ is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 2\\ {2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { \frac{1}{\sqrt{2}}}&{-\frac{1}{\sqrt{2}}}\\ \frac{1}{\sqrt{2}}&{\frac{1}{\sqrt{2}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 2\\ {2} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 0}\\ { 2\sqrt{2}} \end{array}} \right] \end{array}$
c) See image.
