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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

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  • $\begingroup$ Why do you claim that the theorem is not applicable in this case? $\endgroup$ Oct 23 at 13:07
  • $\begingroup$ I think that otherwise I can prove that $AB = I_n$ $\endgroup$
    – Tortar
    Oct 23 at 13:08
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    $\begingroup$ Because the theorem is not true. Left inverse is not always a right inverse. This is true if $T$ is a linear transformation between two finite dimensional spaces of the same dimension. But in general, a function might only have a left inverse. $\endgroup$
    – Mark
    Oct 23 at 13:10
  • $\begingroup$ As an example of a linear transformation that only has one type of inverse but not both, consider the shift operator: $(x_1,x_2,x_3,\ldots)\mapsto(x_2,x_3,\ldots)$. $\endgroup$ Oct 23 at 13:11
  • $\begingroup$ The issue here is that the definition of left inverse given is wrong - or at least highly non-standard. See for example en.wikipedia.org/wiki/Inverse_function#Definitions. That is $S$ is usually defined to be a function $S\colon W\to V$. $\endgroup$
    – tkf
    Oct 23 at 13:23
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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:

  1. If $m > n$ this is not possible by the rank-nullity theorem.
  2. 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 $BA = I_n = I_m$.
  3. 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 $W = \mathbb{F}^n$.
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    $\begingroup$ Let me also emphasize that the definition presented of a left inverse is highly non-standard and while the theorem stated is true, it is better to ignore this definition and use the more standard definition where you do not mess up with the codomains. Most people will be very surprised if you tell them that a map has at most one left inverse. $\endgroup$
    – levap
    Oct 23 at 13:27
  • $\begingroup$ thank you @levap, really well explained! $\endgroup$
    – Tortar
    Oct 23 at 22:57
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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.

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