## 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