# Proving that Matrix is a Unit [duplicate]

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

• That matrix is an absolute unit – DaveBensonPhillips Oct 17 '18 at 22:01
• I don’t understand.... what is your definition of a unit of not the second thing? – rschwieb Oct 17 '18 at 22:07
• @DaveBensonPhillips what’s an absolute unit? – rschwieb Oct 17 '18 at 22:07
• 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. – darij grinberg Oct 17 '18 at 22:09
• 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. – 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 $$ba=1$$