## 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 \(ST = I \in \mathcal{L}(V,V)\).

#### Creating \(S\)

Define a relation \(S:range\ T\to V\) in the following manner:

\begin{equation} S(v) = a \mid Ta = v \end{equation}

#### Demonstrating that \(S\) is a function

We show that there are no two possible choices for \(a\), and therefore that \(S\) is a function, by the injectivity of \(T\). Recall a function is a map has the property that \(v=u \implies Fv=Fu\).

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 \(v=u \implies Ta=Tb\)

Recall now that a linear map called injective when \(Tv=Tu \implies v=u\).

Therefore, from the injectivity of \(T\), we have that \(Ta=Tb \implies a=b\). Hence demonstrating the desired quality that shows \(S\) as a function.

#### Demonstrating that \(S\) is a linear map

The linearity \(S\) actually simply inherits the linearity of \(T\), which is defined to be a linear map.

Additivity:

\begin{align} Sv+Su &= (a \mid Ta =v) + (b \mid Tb =u) \\ &= (a+b) \mid Ta+Tb = (v+u) \\ &= (a+b) \mid T(a+b) = (v+u) \\ &= x \mid Tx = (v+u) \\ &= S(v+u) \end{align}

Homogenity is shown in a similar fashion. We can therefore conclude that \(S \in \mathcal{L}(range\ T, V)\).

#### Note on the codomain of \(S\)

Note that we desire \(S \in \mathcal{L}(W,V),\ i.e.\ S:W\to V\), And yet, as it stands, \(S: range\ T \to W\). Fortunately, as \(range\ T\) is a subspace of \(W\) (as ranges are subspaces of the codomain), we can leverage Axler 3.A-E11 (Sasha’s Proof, “maps to subspaces can be extended to the whole space”) to arbitrary extend \(S\) to \(S:W\to V\).

It turns out that where the “extended” basis vectors gets mapped doesn’t matter. We only care about \(S\) insofar as its compositional behavior with \(T\).

#### Demonstrating that \(S\) has the properties we desire

We desire that \(ST = I \in \mathcal{L}(V,V)\).

Recall that the “identity map” on \(V\) is a map \(I \in \mathcal{L}(V,V)\) such that \(Iv = v, \forall v \in V\). We now show that \(ST\) acts like the identity map.

WLOG take \(v \in V\).

- Let \(Tv=a\).
- Let \(Sa = u\). Based on the definition of \(S\) (that \(Sx = y \mid Ty=x\), “\(S\) is the inverse map”), we have that \(Tu=a\).

Recall once again that a linear map called injective when \(Tv=Tu \implies v=u\).

We now have that \(Tu=a=Tv\), therefore, because \(T\) is given injective, \(u=v\).

We have show WLOG that \((ST)v = S(Tv) =Sa = u=v\). Therefore \((ST)v=v\), making \(ST\) an identity map \(ST:V\to V\). Lastly, as the product of linear maps are themselves a linear map, \(ST=I\in \mathcal{L}(V,V)\)

#### Conclusion

Having constructed the existence of \(S\) based on the required properties of \(T\), we show that given an injective \(T \in \mathcal{L}(V,W)\), have an \(S \in \mathcal{L}(W,V)\) such that \(ST = I \in \mathcal{L}(V,V)\), as desired.

### Given \(S\)

Given some \(T \in \mathcal{L}(V,W)\) and that \(\exists S \in \mathcal{L}(W,V): ST=I \in \mathcal{L}(V,V)\), we desire that \(T\) is injective. Fortunately, we essentially just reverse the logic of the last section in the last part of the proof.

Recall that a linear map called injective when \(Tv=Tu \implies v=u\). Suppose for the sake of contradiction that \(\exists u,v: Tv=Tu\) but \(u\neq v\).

- Let \(Tv=Tu=a\)
- Let \(Sa=b\)

Therefore: \((ST)v=(ST)u=S(a)=b\). Just to reiterate, this means that we have:

- \((ST)v=b\implies Iv=b\)
- \((ST)u=b \implies Iu=b\)

Therefore, we have that \(Iv=Iu\) for distinct \(v,u\), which is absurd. Having reached contradiction, we have that \(Tu=Tv\implies u=v\), reaching the definition of injectivity for \(T\). \(\blacksquare\)