Answer
See below
Work Step by Step
$T^{-1}$ exists
Assume $v_1,v_2 \in V\rightarrow v_1+v_2 \in V\\
(T^{-1}T)(v_1+v_2)=v_1+v_2\\
T^{-1}(T(v_1+v_2))=v_1+v_2\\
T^{-1}(T(v_1)+T(v_2))=v_1+v_2$
Since $T^{-1}$ is an inverse transformation of $T$, we have:
$(T^{-1}T)(v_1)=v_1\\
(T^{-1}T)(v_2)=v_2$
with $v \in V$
then $(T^{-1}(T(v_1)+T(v_2))=(T^{-1}T)(v_1)+(T^{-1}T)(v_2)\\
(T^{-1}(T(v_1)+T(v_2))=(T^{-1}(T(v_1))+(T^{-1}(T(v_2))$
Let $w_1=T(v_1)\\
w_2=T(v_2)$
obtain $T^{-1}(w_1+w_2)=T^{-1}(w_1)+T^{-1}(w_2)$
If we let $c$ is a scalar then we have $(T^{-1}T)(cv_1)=cv_1\\
T^{-1}(T(cv_1))=cv_1\\
T^{-1}(cT(v_1))=cv_1\\
\rightarrow T^{-1}(cT(v_1))=c(T^{-1}T)(v_1)=cT^{-1}(T(v_1))\\
\rightarrow T^{-1}(cw_1)=cT^{-1}(w_1)$
Hence, $T^{-1}$ is a linear transformation.
