Linear Transformation $T_{A}$ Is invertible $\iff$ A Is invertible Let $T_{A}$ be the linear mapping corresponding to the matrix A, and $A \in F^{n*n}$
$T_{A}$ Is invertible $\iff$ there is $T_{A}^{-1}$ so $T_{A} \circ T_{A}^{-1}=I $  $T_{A} \circ T_{A}^{-1}(v)=v$ $\iff$ the matrix $I=A*A^{-1} $ $\iff$ A must be invertible
Is this proof is valid?
 A: If you know that matrices and linear mappings are in one-to-one correspondence, $T_I=\text{id}$ and $T_A\circ T_B=T_{AB}$ for any square matrices $A$ and $B$, then the proof is simple.
If $A$ is invertible, then $T_{A^{-1}}T_A=T_{A^{-1}A}=T_I=\text{id}$ and similarly $T_AT_{A^{-1}}=\text{id}$, so $T_A$ is invertible.
On the other hand, if $T_A$ is invertible, there is a matrix $B$ so that $T_A^{-1}=T_B$.
(This where I use the one-to-one correspondence and the fact that the inverse of a linear mapping is linear.)
Then $T_I=\text{id}=T_A\circ T_B=T_{AB}$, so (again by the correspondence) $AB=I$.
Thus $B=A^{-1}$.
(We have actually proved that $T_A^{-1}=T_{A^{-1}}$.)
About your proof:
I have hard time following your reasoning.
Writing $A^{-1}$ already implies that $A$ is invertible, so the end of the reasoning doesn't make much sense.
Also the role of $v$ is not quite clear.
To show that two things are equivalent, it very often is a good idea to do the two directions separately.
It helps make the presentation clearer, and that is very, very important.
Final words:
I don't mean to be overly harsh.
I thought you would rather hear this here now than from your exam's grader later, not only because you can easily ask for clarification here.
A: The proof is not correct. Hidden in there is the assumption that $(T_A)^{-1} = T_{A^{-1}}$, which you have not proved. 
