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If $A$ is a square matrix and there exists a square matrix $B$ such that $AB =1$, than it is known that $BA=1$. This property is proved with some properties from linear algebra. Although I've never seen it be proved just by structures of matrix multiplication, I couldn't find a counterexample of a set with structures of matrix multiplication but left inverse doesn't imply right inverse.

To be more specific, let $X$ be a set and binary operation $\cdot$ is defined on $X$. If $\cdot$ is associative and $X$ has left and right identity(which will be the same), than does $A \cdot B = 1$ for some $A, B\in X$ implies $B \cdot A = 1$?

If not, what other properties of matrix multiplication should we add to this structure of $(X,\cdot)$ in order to get the property?

coxehj4142
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  • Left and right identity will not necessily be the same. – amsmath Oct 25 '19 at 02:47
  • It is not true in general; see https://math.stackexchange.com/questions/70777/a-ring-element-with-a-left-inverse-but-no-right-inverse – pancini Oct 25 '19 at 02:50
  • @amsmath No, I mean I'd like to know the operation such that if there exists an element which has a left inverse then the element also has a right inverse(Then the two inverses must be same.). – coxehj4142 Oct 25 '19 at 02:52
  • @coxehj4142 Then why do you write something about left and right identities? – amsmath Oct 25 '19 at 02:54
  • It is not enough; a monoid may have elements with one-sided inverses but no inverse on the other side. If $A$ is infinite, the set of all functions $f\colon A\to A$ is a monoid under composition, and if $f$ is surjective but not injective then there exists $g$ such that $f\circ g=\mathrm{id}_A$, but there is no element such that $h\circ f=\mathrm{id}_A$. It really is something special about matrices and the way they act, not about the monoid structure. – Arturo Magidin Oct 25 '19 at 02:56
  • @amsmath I meant there is $e_R$ such that for all $A\in X$, $A\cdot e_R = A$, and there also exists $e_L$ which works in a similar manner. – coxehj4142 Oct 25 '19 at 02:57
  • What you mean is that you have a monoid: A set with a binary operation and a two-sided identity element. – Arturo Magidin Oct 25 '19 at 02:58
  • Oh, thanks for the counterexamples!! I didn't know the structure I made actually had a name. Thanks for letting me know. But also, do you have any idea of what other additional structures needed to obtain the property? (I mean, weaker than group) – coxehj4142 Oct 25 '19 at 03:04
  • @coxehj4142 No, because if you have that property, then you actually have a group. ;-) – amsmath Oct 25 '19 at 03:06
  • @amsmath: No, you don’t. There’s a difference between “every element that has a one-sided inverse is invertible” and “every element is invertible”. The latter gives you a group. The former is a property that the monoid of $n\times n$ matrices (under multiplication) has. – Arturo Magidin Oct 25 '19 at 03:09
  • @amsmath A group needs to have inverse for all elements. I mean if some specific element has a left inverse, then it also has a right inverse. I think it's a weaker condition. – coxehj4142 Oct 25 '19 at 03:09
  • Oh, I see. Now I understand what you want. However, the elements that have this property will form a group of course. – amsmath Oct 25 '19 at 03:10
  • There is nothing specific about the properties of multiplication of matrices that ensures this; every monoid can be embedded as a submonoid of a full transformation semigroup (the set of all functions from a set to itself, under composition), and only when that set is finite do one-sided inverses imply invertibility. It really has to do with the fact that matrices are themselves functions acting on a set, with the property that injective, surjective, and bijective are equivalent for that action. I doubt you’ll find any “nice” property other than “one-sided inverse implies invertible”. – Arturo Magidin Oct 25 '19 at 03:14
  • @ArturoMagidin What will be this finite set for matrices? I'm puzzled. – amsmath Oct 25 '19 at 03:18
  • @ArturoMagidin Do you mean that for every monoid, one-sided inverse implies invertible iff monoid is finite? – coxehj4142 Oct 25 '19 at 03:18
  • @coxehj4142 How could he? The monoid of matrices is not finite. – amsmath Oct 25 '19 at 03:19
  • @ArturoMagidin What do you mean by "injective, surjective, and bijective are equivalent" for matrix action on a set? – coxehj4142 Oct 25 '19 at 03:20
  • @amsmath I think he didn't only mean monoid of matrices, but general monoid. – coxehj4142 Oct 25 '19 at 03:21
  • @coxehj4142 You asked "Do you mean that for every monoid, one-sided inverse implies invertible iff monoid is finite?" But this is obviously false. Counterexample: $n\times n$ matrices. Or just $(\mathbb R,\cdot)$. – amsmath Oct 25 '19 at 03:23
  • @amsmath Well, I guess you're right. I was just wondering what "when that set is finite do one-sided inverses imply invertibility" means. – coxehj4142 Oct 25 '19 at 03:32
  • @coxehj4142: An $n\times n$ matrix acts on a vector space. This action is a function from the vector space to itself. And this function has the property that it is injective if and only if it is surjective if and only if it is bijective. – Arturo Magidin Oct 25 '19 at 03:43
  • @coxehj4142 It means the following: Let $M = (X,\cdot)$ be a monoid. Let $a\in X$ and define $f_a : X\to X$ by $f_a(x) := ax$ for $x\in X$. Then $a\mapsto f_a$ is injective since if $ax = bx$ for all $x\in X$, then $a=b$ follows from taking $x=e$. Hence, $M$ can be seen as a submonoid of the set $T_X$ of functions from $X$ to itself. $T_X$ is called the full transformation semigroup of $X$. Now, he says that one has that one-sided invertibility applies invertibility in $T_X$ only if $X$ is finite. But that doesn't help because we want this property only on $M$. – amsmath Oct 25 '19 at 03:44
  • As to the comment: in a full transformation semigroup (which is the semigroup of all functions from a fixed set $X$ to itself), the invertible elements are the bijective functions, the elements with a left inverse are the injective functions, and the elements with a right inverse are the surjective functions. When $X$ is finite, we are looking at functions $f\colon X\to X$, and so injective is equivalent to bijective is equivalent to surjective. In a full transformation semigroup, one-sided invertibility implies invertibility iff the defining set is finite. – Arturo Magidin Oct 25 '19 at 03:45
  • @ArturoMagidin True. But this doesn't help because our submonoid might be much smaller. – amsmath Oct 25 '19 at 03:46
  • @amsmath: Yes; my point is that it’s not going to be an easy property to state just in terms of the monoid structure in any way that is simpler than “the elements with a left inverse form a group” or some such that is immediately equivalent to “if it has a one-sided inverse then it is invertible”. You can give “simpler” conditions in terms of actions (such as in the full transformation semigroup, “the underlying set is finite”) or in matrices (because you are acting on a finite dimensional vector space). – Arturo Magidin Oct 25 '19 at 03:52
  • @ArturoMagidin The point for matrices is not only that the vector space is finite-dimensional but also that the functions are linear. Both these conditions yield the property. – amsmath Oct 25 '19 at 03:54
  • @amsmath: True; I ellided “linear” because it was the “obvious” action, and this property of linear functions does not hold in infinite dimenisonal spaces. Then again, the property does not hold in finite dimensional spaces for non-linear spaces, so the point is taken. – Arturo Magidin Oct 25 '19 at 03:56

2 Answers2

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Take $X = \{f:\mathbb R\to\mathbb R\}$ equipped with composition $\circ$. Can you think of a function that is surjective but not injective?

amsmath
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If you just consider one operation, that is, if you work with monoids, the answer is no. A counterexample is the bicyclic monoid, which is the quotient of the free monoid on two generators $a$ and $b$ under the relation $ab = 1$.

However, let me answer your last question:

What other properties of matrix multiplication should we add to this structure of $(X,\cdot)$ in order to get the property?

It turns out that the property still holds if you work on a commutative semiring, which is, roughly speaking, a ring without subtraction. For a proof, see [1].

[1] Reutenauer, Christophe; Straubing, Howard. Inversion of matrices over a commutative semiring. J. Algebra 88 (1984), no. 2, 350--360.

J.-E. Pin
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