If $T \colon V \to V$ is a linear operation and $B$ is a basis for $V$, then $T$ is one-to-one $\iff$ $[T]$$B$ is invertible. Moreover if the equivalent statements hold, $[T$$-1$$]$$B$ $=$ $[T]$$-1$$B$.
proof for $\Rightarrow$ :
Let $B=\{b$1$,b$2$,...,b$n$\}$ a basis for $V$. Since $[T(b$$i$$)]$$B$ for $i=1,2,...,n$ are coordinate vectors of $T(b$$i$$)$ with respect to $B$ and $T$ is one to one, the column vectors of $[T]$$B$ form a linear independent set, which implies $[T]$$B$ is invertible.
Is there anything wrong with the above proof? And how should I approach the proof for $\Leftarrow$ ?
Thank you
I don't see anything wrong with your proof. To prove the $\Leftarrow$ statement you can either follow Babelfish's comment or do the following. This is probably longer but I think you might gain a better understanding of some concepts from it.
Let $B = \{b_1,b_2, ...,b_n\}$ be a basis of $V$, $\;T : V \rightarrow V$ the linear transformation defined by the $ n \times n$ matrix $[T_B]$.
For all $v \in V, \exists ! v_1,v_2, ... , v_n \in \mathbb{R}$ such that $v = \sum_{i=1}^n v_i b_i$. By uniqueness of the $v_i$'s the following function is well defined.
$$ v \mapsto T(v)= T(\sum_{i = 1}^n v_i b_i) = \sum_{i = 1}^n ([T_B] \cdot v^B)_i b_i$$ where $v^B = (v_1,v_2,...,v_n ) \in \mathbb{R}^n$ are the coordinates of $v$ in the basis $B$. If we denote $T(v)^B$ the coordinates of $T(v)$ in $B$ then the above line is just :
$$ T(v)^B = [T_B]v^B.$$
To show that the function $T$ is one to one it is sufficient to show that it is invertible. Since we know that the matrix $[T_B]$ is invertible let's use it's inverse to explicitly construct the inverse of $T$.
Define $T': V \rightarrow V$ by using $[T_B]^{-1}$ i.e. $$ v \mapsto T'(v) = \sum_{i = 1}^n ([T_B]^{-1} \cdot v^B)_i b_i.$$
Now we show that this is the inverse of $T$
\begin{align*} T'(T(v)) &= \sum_{i = 1}^n ([T_B]^{-1} \cdot T(v)^B)_i b_i \\ &= \sum_{i = 1}^n ([T_B]^{-1} \cdot ([T_B] \cdot v^B))_i b_i \\ &= \sum_{i = 1}^n ([T_B]^{-1} \cdot [T_B]) \cdot v^B)_i b_i \\ &= \sum_{i = 1}^n (v^B)_i b_i \\ &= \sum_{i = 1}^n v_i b_i \\ &= v \end{align*}
This shows us that $T'$ is an inverse to the left of $T$. It can also be shown by similar arguments that $T'$ is an inverse to the right of $T$.
Also notice that you can use the same method to prove $\Rightarrow.$
by calling $[T_B]^{-1}$ the matrix of $T^{-1}$ then by using the same calculations you would get
$$v = T^{-1} (T(v)) = \sum_{i = 1}^n ([T_B]^{-1} \cdot [T_B]) \cdot v^B)_i b_i $$
but $v = \sum_{i=1}^n v_i b_i$ and since $B$ is a basis you can deduce $$ v_i = (v^B)_i = ([T_B]^{-1} \cdot [T_B]) \cdot v^B)_i \quad \forall v^B = (v_1,...,v_n).$$