Answer
onto, but not one to one
Work Step by Step
$A=[T(e_1)\ \ T(e_2)\ \ T(e_3)]=\begin{bmatrix}
1&4&-5\\
3&-7&4
\end{bmatrix}$~$\begin{bmatrix}
1&0&-1\\
0&1&-1
\end{bmatrix}$
Because A has a pivot in each row, the linear transformation is onto, but because it does not have a pivot in each column, it is not one to one.
