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title: "NUS-MATH530 3.B Problem 20"
source: https://www.jemoka.com/posts/kbhnus_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 \(ST = I \in \mathcal{L}(V,V)\).
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 &amp;= (a \mid Ta =v) + (b \mid Tb =u) \\ &amp;= (a+b) \mid Ta+Tb = (v+u) \\ &amp;= (a+b) \mid T(a+b) = (v+u) \\ &amp;= x \mid Tx = (v+u) \\ &amp;= 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&rsquo;s Proof, &ldquo;maps to subspaces can be extended to the whole space&rdquo;) to arbitrary extend \(S\) to \(S:W\to V\).
It turns out that where the &ldquo;extended&rdquo; basis vectors gets mapped doesn&rsquo;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 &ldquo;identity map&rdquo; 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 \(Sx = y \mid Ty=x\), &ldquo;\(S\) is the inverse map&rdquo;), 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 \(T\). \(\blacksquare\)
