How to prove that if $U$ and $V$ are finite dimensional vector spaces and $T:U \rightarrow V$ is a linear transformation, then $T$ is injective if and only if $S \circ T$ is the identity function on $U$ for some linear transformation $S$?
[Math] Linear transformation is injective if and only if there exists a linear transformation where the composition is identity
linear algebra
Best Answer
Suppose $U$ and $V$ are finite-dimensional vector spaces and $T:U \rightarrow V$ is a linear transformation. Define $R: range(T) \rightarrow V$ with $R(T(u))=u$; this can be done because $T$ is injective, that is, for each $T(u) \in range(T)$ for some $u \in U$, we have $T^{-1}(T(u))=\{u\}$, a singleton. Now, for $u,u' \in U$, and $\alpha$ a scalar, $$R[\alpha T(u)+T(u')]=R[T(\alpha u+u')]=\alpha u+u'=\alpha R(T(u))+R(T(u'));$$ $R$ is then a linear transformation and there exists an extension of $R$, a linear transformation $S:V \rightarrow U$ that satisfies $(ST)(u)=S(T(u))=R(T(u))=u$ whenever $u \in U$. Therefore $S \circ T$ is the identity on $U$. Conversely, suppose there exists a linear transformation $S:V \rightarrow U$, which satisfies $(ST)(u)=u$ whenever $u \in U$. If $T(u) \in range(T)$ for some $u \in U$, and if $u' \in T^{-1}(T(u))$, then $u'=S(T(u'))=S(T(u))=u$. Therefore $T$ is injective.