Answer
See below
Work Step by Step
Assume $T$ is one-to-one
Obtain: $Ker(T)=\{0\} $
then $\dim[Ker(T)]=0$
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim V\\
0+\dim [Rng(T)]=\dim V\\
\dim [Rng(T)]=\dim V$
Since $Rng(T) \subset V$ and $\dim W=\dim [Rng(T)]$ we have $Rng(T)=W$
Hence, $T$ is onto
Assume $T$ is onto
Since $Rng(T)=W$ we have $\dim [Rng(T)]=\dim W\\
\rightarrow \dim [Rng(T)]=\dim V$
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim V\\
\dim [Ker(T)]+\dim V=\dim V\\
\dim [Ker(T)]=0\\
Ker(T)=\{0\}$
Hence, $T$ is one-to-one.
