If $T: V\rightarrow W$ is isomorphic then $T$ is invertible. Prove that its inverse is a linear transformation Ok so I know $T$ is a linear transformation so it stands that this is true: 
$T(\alpha x + y) = \alpha T(x) + T(y)$
However, I'm a little lost on how I should use this information to prove that the inverse is also a linear transformation. 
 A: Suppose that T is invertible so we have $T = (T^{-1})^{-1}$ I will denote $T^{-1} = M$ just for clarification. Note that $TT^{-1} = I$
So since $T = M^{-1}$ $M(x + y) = M(T(M(x)) + T(M(y))$ now Since T is 
linear transformation we have $M(T(M(x)) + T(M(y))) = M(T(M(x) + M(y)))$
Now since M and T are inverse with each other they will cancel to identity
and we will have $M(T(M(x) + M(y))) = M(x) + M(y)$ 
so we have $T^{-1}(x + b) = M(x + y) = M(x) + M(y) = T^{-1}(x) + T^{-1}(y)$
and you are done for the sake of completeness here is also the second property of linear transformation. 
$T^{-1}(ax) = T^{-1}(aT(T^{-1}(x))) = T^{-1}(T(aT^{-1}(x))) = T^{-1}(aTT^{-1}(x)) = aT^{-1}(x)$.
A: If $T: V \to W$ is an isomorphism, then for any $w_1, w_2 \in W$ there exist unique $v_1, v_2 \in V$ with
$$w_1 = T(v_1) \quad\text{and}\quad w_2 = T(v_2).$$
(Why do they exist, and why are the unique?)
Now, can you show $T^{-1}(\alpha w_1 + \beta w_2) = \alpha T^{-1}(w_1) + \beta T^{-1}(w_2)$, using only the linearity of $T$? (Remember, $v_1 = T^{-1}(w_1)$, similarly for $v_2$).
