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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.


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

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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