Why can’t I apply theorem on existence and uniqueness of inverses in this case? This is a follow up of my last question where I was not clear on my problem.
What I want to ask is :
Given a $n \times m$ matrix $A$ and a $m \times n$ matrix $B$ such that $$BA = I_m$$
why $AB$ is always different from $I_n$ if $ n \neq m$ ? And specifically why is the theorem

A function $T : V \to W$ can have at most one left inverse. If $T$ has a left inverse $S$, then $S$ is also a right inverse.

not applicable in this case?
Given that in the book these definitions are given at the start of the section on inverses :

Given two sets $V$ and $W$ and a function $T : V \to W$. A function $S : T(V) \to V $ is called a left inverse of $T$ if $S[T(x)] = x$ for all $x$ in $V$, that is, if $$ST = I_V$$ where $I_V$ is the identity transformation on $V$.


A function $R : T(V) \to V$ is called a right inverse of $T$ if $T[R(y)] = y$ for all $y$ in $T(v)$, that is, if $$TR= I_{T(V)}$$ where $I_{T(V)}$ is the identity transformation on $T(V)$.

I’m not asking how to prove that $n =m$ is necessary to have a left inverse equal to a right inverse, but why the theorem I quoted is not applicable. Thank you in advance
 A: Set $V = \mathbb{F}^m, W = \mathbb{F}^n$ and consider the linear maps $T = T_A \colon \mathbb{F}^m \rightarrow \mathbb{F}^n$ and $S = S_B \colon \mathbb{F^n} \rightarrow \mathbb{F}^m$ given by
$$ T(\vec{x}) = T_A(\vec{x}) = A\vec{x}, \qquad S(\vec{y}) = S_B(\vec{y}) = B\vec{y}. $$
The identity $BA = I_m$ translates into the identity $S \circ T = S_B \circ T_A = I_{\mathbb{F}^m}$.
Let's try to apply the theorem to this case. Note that in the theorem $S$ is a function from $T(V)$ to $V$ while our $S$ is a function from $W$ to $V$. To be in the situation of the theorem, we need to restrict $S$ to $T(V)$ and then $S|_{T(V)} \circ T = I_V$. Then we also have $T \circ S|_{T(V)} = I_{T(V)}$.
However, this does not say that $AB = I_n$! The only thing this says is that if $\vec{y}$ is in the image of $A$ then $AB\vec{y} = \vec{y}$. However, in order for $AB = I_n$ to hold, you need $AB \vec{y} = \vec{y}$ to hold for all $\vec{y} \in \mathbb{F}^n$.
In fact, the situation is as follows. The identity $BA = I_m$ (or the equivalent identity $S \circ T = I_{\mathbb{F}^m}$) implies that $T$ is one-to-one. Then:

*

*If $m > n$ this is not possible by the rank-nullity theorem.

*If $m = n$ then by the rank-nulity theorem $T = T_A$ is also onto so $T(V) = W$ and then the uniqueness result indeed gives you $T \circ S|_{T(V)} = T \circ S = I_{W}$ so $BA = I_n = I_m$.

*If $m < n$ then $T = T_A$ will never be onto so the "identity" $T \circ S = I_{\mathbb{F}^n}$ ony holds on $T(V)$ which is strictly smaller than $W = \mathbb{F}^n$.

A: As stated, the theorem is correct, but it uses an unorthodox definition of left inverse. Here, if $T$ is a map from a set $V$ into a set $W$, then a left inverse is defined as a map from $T(V)$ into $V$, and not, as it is usual, as a map from $W$ into $V$. If we define a left inverse of $T$ as a map $S\colon W\longrightarrow V$ such that $S\circ T=\operatorname{Id}$ (the usual definition) then the theorem doesn't hold anymore and so you cannot apply it.
