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This question already has an answer here:

If $A \in M_{n}(F)$, I have to show that $A$ is a unit only if $AB = I$ or $BA = I$ for some $B \in M_{n}(F)$.

I am not sure how to approach this at all since this fact was pretty intuitive to me. One way I am thinking this is using the fact that this basically means prove that $A$ is invertible if $AB=BA=I$. A hint that was given, however, was that if $V$ and $W$ are finite dimensional vector spaces, $T: V \to W$ is isomorphic only when it is injective (or only when it is surjective). This hint is completely throwing me off on how to do this question. Can somebody help? Thanks.

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marked as duplicate by André 3000, Community Oct 18 '18 at 1:56

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    $\begingroup$ That matrix is an absolute unit $\endgroup$ – DaveBensonPhillips Oct 17 '18 at 22:01
  • $\begingroup$ I don’t understand.... what is your definition of a unit of not the second thing? $\endgroup$ – rschwieb Oct 17 '18 at 22:07
  • $\begingroup$ @DaveBensonPhillips what’s an absolute unit? $\endgroup$ – rschwieb Oct 17 '18 at 22:07
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    $\begingroup$ You need to prove that if you have $AB = I$ or $BA = I$, then you have $AB = BA = I$. This is what the hint can help you with. $\endgroup$ – darij grinberg Oct 17 '18 at 22:09
  • $\begingroup$ Recall that matrix multiplication corresponds to a composition of linear transformations. Since the identity is an isomorphism, and AB = I, that says that AB is injective and surjective. What can be said about A and B separately if AB is injective and surjective? Note also that the same thing can be said for BA = I, so anything that is true for A will be true for B. $\endgroup$ – Joel Pereira Oct 17 '18 at 22:11
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A ring $R$ (with unity) is called Dedekind-finite if for all $x,y \in R$ we have $xy = 0 \to yx = 0$.

I agree with @darijgrinberg - the only way I can understand the OP's question is to assume it is a slightly mis-phrased version of the following:

Show $A \in M_{n}(F)$ is a unit if we are only given that there is a $B \in M_{n}(F)$ for which either $AB=I$ or $BA = I$.

We further presume, from a convention of notation, and the mention of vector spaces in the hint, that $F$ is a field. So OP's question (on this interpretation) can be rephrased in a more lucid fashion as:

Show that $ M_{n}(F)$ is a Dedekind-finite ring.

The $n$-dimensional matrix algebra $M_n(F)$ is isomorphic to the algebra of linear transformations on the vector space $F^n$. Thus multiplication of elements is interpreted as the composition of two linear transformations.

This gives us the three crucial properties, from which the Dedekind-finite property can be inferred:

(a) every element can be uniquely assigned a rank, which is an integer between $0$ and $n$. This rank is in fact equal to the dimension of the image when the linear transformation corresponding to the element (wrt some pre-chosen basis for $F^n$) is applied to the whole of $F^n$

(b) an element is a unit (invertible) if and only if it has maximal rank ($n$)

(c) the rank function $\rho$ satisfies: $$ \forall X,Y \in M_n(F) \quad \quad \rho(XY) \le \max(\rho(X),\rho(Y)) $$

Note 1: in case the field $F$ is finite, then the Dedekind-finiteness of $M_n(F)$ is a consequence of an ingenious theorem of Kaplansky that if an element $a$ of a ring has a left (right) inverse, then either $a$ is a unit or it has an infinite number of distinct left(right) inverses.

Note 2: the Dedekind-finiteness of $M_n(F)$ also follows from a more general result, that any Noetherian ring is Dedekind-finite. To show this, suppose we are given that $ab = 1$.

the map $L_a:R \to R$ defined by $L_a(x) = ax$ is obviously linear. it is also surjective, since for any $r \in R$ we have $L_a(br) = abr = r$. Let $K_n$ be the kernel of $L_a^n$. Obviously the $K_n$ are ideals and: $K_1 \subset K_2 \subset\dots$. The Noetherian condition shows that this chain stabilizes after a finite number of steps, say in $K_n$. Since $L_a^n$ is surjective there must be an $x$ such that $L_a^n(x)=ba-1$. but clearly $L_a(ba-1) = 0$ so $x \in K_{n+1} = K_n$, so $ba-1 = L_n(x) = 0$ and $ba=1$

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