I'm working with a non-commutative, non-associative, non-unital algebra with objects that are somewhat like matrices, but have the property that for almost all $X$ (except the "zero-like" elements), there exists a unique element $X'$ such that for all $A$, $X'AX = XAX' = A$ and $X'' = X$. However, there is no similar element $X^*$ such that for all $A$, $(AX)X^* = A$.

I know there are "right-inverses" and "left-inverses", but is there a name for when the inverse element has to be on the opposite side in order to undo the operation?

Also, I'm referring to inverse elements which isn't strictly accurate according to Wikipedia, but really, these are just elements which reverse multiplication by $X$, but only if on the opposite side of the operation.

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    $\begingroup$ I would strongly advice you to use $X’$ instead of $X^{-1}$; as you can see from the answers, people are confusing this with an actual inverse. What you have is that given $A$, for almost all $X$ there exists a unique $X’$ such that $XAX’ = A$ and $X’AX=A$. $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:03
  • $\begingroup$ Does the $X^{-1}$ you have depend only on $X$, or on both $X$ and $A$? If the latter, you should definitely call it something that reflects the dependence, such as $f(X,A)$ or something like that.... $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:05
  • $\begingroup$ (No, I don’t think there is a name for such an element; it is vaguely reminiscent of the formulas for pseuoinverses, but not quite right for that either) $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:05
  • $\begingroup$ That's a good point, I've changed the notation and added a note. $\endgroup$ – bjshnog Jul 26 '19 at 4:08
  • $\begingroup$ $X'$ can be calculated directly from $X$. $\endgroup$ – bjshnog Jul 26 '19 at 4:08

You should take a look at the commutator.

Depending on the definition that you use, you essentially have that almost all A have a corresponding element X such that the commutator [A,X] is the identity element.

$[A,X] = AXA^{-1}X^{-1}$

(Sorry about formatting I typed this on my phone)

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  • $\begingroup$ I've updated my post to clarify that my algebra does not have identity. $\endgroup$ – bjshnog Jul 26 '19 at 4:01

If $e$ is an identity element of (S,*) (i.e., S is a unital magma) and $a \times b = e$, then $a$ is called a left inverse of $b$ and $b$ is called a right inverse of $a$. If an element $x$ is both a left inverse and a right inverse of $y$, then $x$ is called a two-sided inverse, or simply an inverse, of $y$.

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  • $\begingroup$ I should clarify, there are no identity elements in this algebra, but multiplication can still be reversed by multiplying with a unique element. $\endgroup$ – bjshnog Jul 26 '19 at 3:56
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    $\begingroup$ Is your $a\ x\ b = e$ meant to be $a\times b=e$? Because you use $x$ later to represent an element; and if they are three elements, then $axb=e$ does not yield that $a$ is a left inverse of $b$ and $b$ a right inverse of $a$. Use \times for $\times$; do not use x. $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:00
  • $\begingroup$ Also, what the OP has is that given $a$, for every $b$ there exists a unique $b’$ such that $bab’ = a$ and $b’ab=a$. $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:01
  • $\begingroup$ This does not actually answer the question being posed. $\endgroup$ – Arturo Magidin Jul 26 '19 at 4:07
  • $\begingroup$ thank you everyone for pointing out my mistakes, i will try to change my answer $\endgroup$ – user689080 Jul 26 '19 at 5:59

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