NUS-MATH530 3.B Problem 20

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 “identity map” 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