---
title: "NUS-MATH530 3.B Problem 20"
source: https://www.jemoka.com/posts/kbhnus_math530_3_b_problem_20-1/
---

Statement Support \(W\) is finite-dimensional, and \(T \in \mathcal{L}(V,W)\). Prove that \(T\) is injective IFF \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).
Proof Given injectivity Given an injective \(T \in \mathcal{L}(V,W)\), we desire that \(\exists S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\).
We begin with some statements.
Recall that, a linear map called injective when \(Tv=Tu \implies v=u\) Recall also that the &ldquo;identity map&rdquo; on \(V\) is a map \(I \in \mathcal{L}(V,V)\) such that \(Iv = v, \forall v \in V\) Motivating \(S\) We show that we can indeed create a function \(S\) by the injectivity of \(T\). Recall a function is a map has the property that \(v=u \implies Fv=Fu\).
WLOG consider two vectors \(a,b \in V\).
Creating \(S\) Define a function \(S:W\to V\) in the following manner:
\begin{equation} S(v) = a \mid Ta = v \end{equation}
Demonstrating that \(S\) is a function
So, given \(v, u \in W\) and \(v=u\), we have:
\(Sv = a \mid Ta=v\) \(Su = b \mid Tb=u\) If \(Sv=Su\), then \(a=b\). To demonstrate that \(S\) is a function, we now desire that \(a=b\).
From the above, we have that \(Ta=v\), \(Tb=u\). From prior, we have \(v=u\) From the two statements above, we have \(v=u \implies Ta=Tb\) Lastly, from the injectivity of \(T\), we have that \(Ta=Tb \implies a=b\) Hence demonstrating
