Answer
See below
Work Step by Step
Given that $T$ is a one-to-one linear transformation (1), we have $dim [Ker(T)]=0s$
According to Rank- Nullity Theorem:
$dim[Ker(T)]+dim[Rng(T)]=dim V \\
0+dim[Rng(T)]=dim V\\
dim[Rng(T)]=dimV$
Since $Rng(T) \subset V\\
dim[Rng(T)]=dimV\\
\rightarrow Rng(T)=V$
Therefore,$ T$ is onto (2)
From (1) and (2), $T^{-1}$ exists
