Answer
a) The matrix corresponding to the 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) The image of
$v=\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right]$
is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 3}\\ { - 4} \end{array}} \right] \end{array}$
c) See image.
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 the 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
$v=\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right]$
is
$\begin{array}{l} T(v) = Av\\ \Rightarrow T\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 1}&0\\ 0&{ - 1} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 3\\ 4 \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - 3}\\ { - 4} \end{array}} \right] \end{array}$
c) See below.
