Answer
$T$ is not a linear transformation.
Work Step by Step
$T$ is not a linear transformation.
Take $n=2$
$\begin{array}{l}
T\left( {2\left[ {\begin{array}{*{20}{c}}
1&0\\
0&2
\end{array}} \right]} \right) = T\left( {\left[ {\begin{array}{*{20}{c}}
2&0\\
0&4
\end{array}} \right]} \right) = \frac{1}{8}\left[ {\begin{array}{*{20}{c}}
4&0\\
0&2
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{1/2}&0\\
0&{1/4}
\end{array}} \right]\\
2T\left( {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&2
\end{array}} \right]} \right) = 2\left( {\frac{1}{2}\left[ {\begin{array}{*{20}{c}}
2&0\\
0&1
\end{array}} \right]} \right) = \left[ {\begin{array}{*{20}{c}}
2&0\\
0&1
\end{array}} \right]
\end{array}$
Since, $T\left( {2\left[ {\begin{array}{*{20}{c}}
1&0\\
0&2
\end{array}} \right]} \right) \ne 2T\left( {\left[ {\begin{array}{*{20}{c}}
1&0\\
0&2
\end{array}} \right]} \right)$
Thus, $T$ is not a linear transformation.
