Answer
$ \left[ {\begin{array}{*{20}{c}}
{ - 2}\\
4\\
2\\
0
\end{array}} \right]$
Work Step by Step
Take the standard basis for $R^4$, which is
$(1,0,0,0),(0,1,0,0,),(0,0,1,0),
(0,0,0,1)=\{e_1,e_2,e_3,e_4\}$.
Given
$\displaystyle
T(x_1,x_2,x_3,x_4)=(x_1-x_3,x_2-x_4,x_3-x_1,x_2+x_4)$
Then,
$T(1,0,0,0)=(1-0,0-0,0-1,0+0)=(1,0,-1,0)\\
T(0,1,0,0)=(0-0,1-0,0-0,1+0)=(0,1,0,1)\\
T(0,0,1,0)=(0-1,0-0,1-0,0+0)=(-1,0,1,0)\\
T(0,0,0,1)=(0-0,0-1,0-0,0+1)=(0,-1,0,1)$
Therefore, the matrix corresponding to the linear transformation $T$ is given by:
$A=[T(e_1)$ $T(e_2)$ $T(e_3)$ $T(e_4)]$
$=\left[ {\begin{array}{*{20}{c}}
1&0&{ - 1}&0\\
0&1&0&{-1}\\
{-1}&0&1&0\\
0&1&0&1
\end{array}} \right]$
Hence, the image of
$v=\left[ {\begin{array}{*{20}{c}}
1\\
{ 2}\\3\\{-2}
\end{array}} \right]$
is
$\begin{array}{l}
T(v) = Av\\
\Rightarrow T\left[ {\begin{array}{*{20}{c}}
1\\
2\\
3\\
{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
1&0&{ - 1}&0\\
0&1&0&{ - 1}\\
{ - 1}&0&1&0\\
0&1&0&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
1\\
2\\
3\\
{ - 2}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{ - 2}\\
4\\
2\\
0
\end{array}} \right]
\end{array}$
