Answer
a) The matrix corresponding to linear transformation $T$ is given by: $=\left[ {\begin{array}{*{20}{c}} { 1}&{0}\\ 0&{- 1} \end{array}} \right]$
b) The image of $v=\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right]$ is $\left[ {\begin{array}{*{20}{c}} { 4}\\ { 1} \end{array}} \right] $
c) See image below.
Work Step by Step
a) Take the standard basis for $R^2,$ which is $\{(1,0),(0,1)\}=\{e_1,e_2\}$.
Given
$T(x,y)=(x,-y)$
Then,
$T(1,0)=(1,0)\\ T(0,1)=(0,-1)$
Therefore, the matrix corresponding to linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)]$ $=\left[ {\begin{array}{*{20}{c}} { 1}&{0}\\ 0&{- 1} \end{array}} \right]$
b) Hence, the image of $v=\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right]$ is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 1}&0\\ 0&{-1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 4\\ {-1} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { 4}\\ { 1} \end{array}} \right] \end{array}$
c)
