Proving that a composition is a linear transformation

Question

Let $T: U \rightarrow V$ be an isomorphism.

Given that $S: V \rightarrow U$ satisfies $T \circ S = Id_V$ (where $Id_V (v) = v$, for all $v \in V$), show that $S$ is a linear transformation.

I honestly have no clue in solving this exercise. I would really appreciate a hint. How can I even start?

Answer 1

Observe that $S$ is in fact the inverse transformation of $T$. If this does not appear completely obvious, we can show it like this: Observe that since $T$ is an isomorphism, it is invertible. We apply it to the equation from the question and get

$$T^{-1} \circ (T \circ S) = T^{-1} \circ \text{id}_V$$

$$(T^{-1} \circ T) \circ S = T^{-1}$$

$$ \text{id}_V \circ S = S = T^{-1}$$

Now, since $T$ is an isomorphism, its inverse is so as well, and in particular it is linear. You can also show this from the definition, if you want to.

Answer 2

Let $v_1,v_2\in V$. Then, for any scalar $a$, \begin{align} T(S(av_1+v_2)) &= av_1+v_2 \\ &= aT(S(v_1)) + T(S(v_2)) \\ &= T(aS(v_1)+S(v_2)) \qquad \textrm{since $T$ is linear} \end{align} and then, injectivity implies that $S(av_1+v_2) = aS(v_1)+S(v_2)$, that is, $S$ is linear.