Answer
$T$ is a linear transformation.
Work Step by Step
A linear transformation is said to be linear if $T(av+w)=aT(v)+T(w)$, for all $v$,$w$ in the given domain and for all scalars $a$.
Here the domain is $M_{2,2}$. Therefore, let
$v = \left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right],w = \left[ {\begin{array}{*{20}{c}}
e&f\\
g&h
\end{array}} \right]$
be arbitrary elements of $M_{2,2}$ and $k$ be any scalar.
Then,
$\begin{array}{l}
T(kv + w)\\
= T\left( {k\left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
e&f\\
g&h
\end{array}} \right]} \right)\\
= T\left( {\left[ {\begin{array}{*{20}{c}}
{ka}&{kb}\\
{kc}&{kd}
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
e&f\\
g&h
\end{array}} \right]} \right)\\
= T\left( {\left[ {\begin{array}{*{20}{c}}
{ka + e}&{kb + f}\\
{kc + g}&{kd + h}
\end{array}} \right]} \right)\\
= (ka + e) + (kb + f) + (kc + g) + (kd + h)\\
= (ka + kb + kc + kd) + (e + f + g + h)\\
= k(a + b + c + d) + (e + f + g + h)\\
= kT\left( {\left[ {\begin{array}{*{20}{c}}
a&b\\
c&d
\end{array}} \right]} \right) + T\left( {\left[ {\begin{array}{*{20}{c}}
e&f\\
g&h
\end{array}} \right]} \right)\\
= kT(v) + T(w)
\end{array}$
Since, $k$,$v$ and $w$ are arbitrary, hence,
$T(kv+w)=kT(v)+T(w)$ for all scalars $k$ and for all $v,w$ in $M_{2,2}$.
Therefore, $T$ is a linear transformation.
