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 \(T\). \(\blacksquare\)