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In the accepted answer of this question it is stated in the comments section that $CA$ is not a right inverse of $B$ (even if $CA$ is a left inverse of $B$) when $A$ and $B$ are not square matrices.

But why is this the case? Isn't always the case that if you have a left inverse then it is always a right inverse?

I read in In Apostol's Calculus II this theorem which causes me this confusion :

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.

I'm surely not understanding the connection between functions and matrices, can someone clarify?

I'm not referring to inverses in a strict sense. Say for example $A$ is $ n \times m$ and $B$ is $m \times n$, if $AB = I_n$ and $BA = I_m$ for me it is fine and I think also in the comment section of the answer it was understood this way.

In the book It is given this definition 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 $T(V)$.

After this definition and an example It is stated the theorem I mentioned, is it wrong?

Tortar
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    Nonsquare matrices can’t have inverses. – Randall Oct 22 '21 at 21:40
  • I'm not referring to inverses in a strict sense. Say for example $A$ is $ n \times m$ and $B$ is $m \times n$, if $AB = I_n$ and $BA = I_m$ for me it is fine – Tortar Oct 22 '21 at 21:44
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    It's not possible for $AB = I_n$ and $BA = I_m$ unless $n = m$. The statement you quote looks very suspicious - can you provide more context for the quote. – Rob Arthan Oct 22 '21 at 21:47
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    Recall that the rank of $AB$ is at most the rank of $A$ and is at most the rank of $B$. So if $m<n$, it's possible for $BA$ to equal $I_m$, but it's not possible for $AB$ to equal $I_n$. — As for the theorem you quoted from your book, without additional assumptions it's definitely false as stated (do you mean $T\colon V\to V$ instead of $V\to W$?). – Greg Martin Oct 22 '21 at 21:47
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    @Randall: the question is about left inverses and right inverses. Non-square matrices can have left or right inverses (but not both). – Rob Arthan Oct 22 '21 at 21:49
  • @GreG Martin the $S$ is defined on $T(V) \to V$, does it change something? The theorem was as I stated it : theorem 2.8 of Apostol Calculus II – Tortar Oct 22 '21 at 21:52
  • I no longer possess Apostol's books. Please state the hypotheses on the vector spaces $V$ and $W$ — these are clearly missing from your quotation. – Ted Shifrin Oct 22 '21 at 21:55
  • @Ted Shifrin, I added the relevant definitions on the inverses, this is all is specified for $V$ and $W$. – Tortar Oct 22 '21 at 22:04
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    I found a PDF of Apostol's book online. His notion of left and right inverse is non-standard. For most of us a (left or right) inverse to a function $T : V \to W$ is a function $W \to V$. Apostol says it's a function from the range $T(V)$ to $V$,. Hence the facts Apostol states about functions and corresponding facts about matrices that represent linear transformations have a strange feel. – Rob Arthan Oct 22 '21 at 22:07
  • thanks @Greg Martin to explain why it is true, but anyway I'm still confused by the theorem I quoted – Tortar Oct 22 '21 at 22:20
  • thanks @Rob Arthan, could you try to help me with this confusion? Is the theorem related to the question I asked ? Why is it not in contradiction with the fact that in this case a left inverse is not the same as the right? – Tortar Oct 22 '21 at 22:22
  • the requirement $S(T(x)) = x$ completely specifies what $S$ must to do to elements $y \in T(V)$. If $y \in W \setminus T(V)$, then $S$ can do what it likes to $y$ without affection the truth of $S(T(x)) = x$. Try working it through with some simple examples like $T : \Bbb{N} \to \Bbb{N}$ defined by $T(n) = \mathrm{max}(n, 10)$. – Rob Arthan Oct 22 '21 at 22:29
  • I think that $S(1)$ in your example can be whatever even if $S$ itself doesn't exist actually because $T$ is not one-to-one, but still I don't see the connection with inverses of matrices and the restriction to the range of $T$ unfortunately – Tortar Oct 22 '21 at 23:38

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Let $A\in\mathbb R^{n\times m}$ and $B\in\mathbb R^{m\times n}$ such that $AB\in\mathbb R^{n\times n}$ is invertible and $m\neq n$. Then $n<m$. Indeed, if $n>m$, then the rank-nullity formula applied to $B$ gives $$ n = \dim\operatorname{im}B + \dim\ker B\le m + \dim\ker B, $$ and hence $\dim\ker B\ge n-m>0$. But $\ker B\subset\ker(AB)$ and thus $\ker(AB)\neq\{0\}$. But $AB$ is invertible, contradiction.

As a consquence, $A$ is flat and $B$ is tall. Moreover, $B$ is injective ($\Leftrightarrow$ has a left inverse $\Leftrightarrow$ $\ker B = \{0\}$) and $A$ is surjective (has a right inverse). But $B$ doesn't have a right inverse. If it had, there would be an $n\times m$ matrix $X$ such that $BX = I_m$, which implies that $\operatorname{im}B = \mathbb R^m$. But that's not possible due to the rank-nullity formula again: $$ n = \dim\operatorname{im}B + \dim\ker B = \dim\operatorname{im}B. $$ Similarly, one shows that $A$ doesn't have a left inverse.

amsmath
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