Answer
$T$ is not a linear transformation.
Work Step by Step
A 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 above condition is violated.
Take, $a=1;v=w=\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]$
Then,
$T\left( {1\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}
2&2\\
2&2
\end{array}} \right]} \right) = {2^2} = 4$
and
$\begin{array}{l}
1T\left( {\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right) + T\left( {\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right) = {1^2} + {1^2} = 1 + 1 = 2\\
\therefore T\left( {1\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right) = 4 \ne 2 = 1T\left( {\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right) + T\left( {\left[ {\begin{array}{*{20}{c}}
1&1\\
1&1
\end{array}} \right]} \right)
\end{array}$
Hence, $T$ is not a linear transformation.