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).
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