# AB = I <--> BA = I, proof not valid?

Ok so I'm right at the end of the first chapter in my book which means I just know some matrix/vector operations, what it means for a matrix to be invertible and what the transpose of a matrix is. Add to that some of the basic rules regarding the inverse and transpose of a matrix, such as $(AB)^{-1} = B^{-1}A^{-1}$

One questions asks me to prove
$AB = I \iff BA = I$
I googled but all I could find was proofs that used concepts I've never heard of, like determinants. The reason I'm asking for help is because I think the book's answer is wrong. Here it is for $AB = I \implies BA = I$:

$B = BI$
$B(AB) = (BA)B$
Post-multiply this $B = (BA)B$ by $B^{-1}$:
$BB^{-1} = (BA)BB^{-1}$
$I = (BA)I = BA$
We have $BA = I$

He proves $BA = I \implies AB = I$ in a similar way. The thing I am turning against is that it is assumed that $B$ is invertible, that $B^{-1}$ exists. The question never said anything about $B$ being invertible. But if $B^{-1}$ actually exists then no proof is needed since then by the definition of an invertible matrix, $A$ is the inverse of $B$ and $AB = BA = I$. So am I wrong or is the proof in the book not really correct? Or is it a valid assumption that $B$ is invertible?

• The book is wrong to assume $B$ is invertible. Unless they have already proven that square matrices have the property that injective or surjective implies invertible, in which case that's valid, but there's not enough information in the part you quoted from to tell. Feb 6, 2015 at 8:41
• Why does AB has to be square? Feb 6, 2015 at 8:47
• Well you quoted a property of inverses $(AB)^{-1}=B^{-1}A^{-1}$, this is only true if the matrices are invertible, which is impossible for non-square matrices. Feb 6, 2015 at 8:59
• Oh I am sorry if I wasn't clear enough, that quote had nothing to do with the question, I just wanted to show which level I am at. So the question should concern matrices of all sizes as long as matrix multiplication is valid. Feb 6, 2015 at 9:03
• Duplicate: math.stackexchange.com/q/3852/472818 Jan 31, 2023 at 18:26

There are thre options here: either you are sloppy or the book is or both.

It is pretty easy to show the following fact:

If $A$ and $B$ are square matrices, then $AB$ is invertible if and only if $A$ and $B$ are invertible.

Using this fact, the proof used in your book is "correct".

1. If the statement above was already proven in your book, then it may be the case that you missed that fact (so you are sloppy)
2. If the statement was not proven in the book, then the book is sloppy.
3. If the statement was proven, but the book did not explain well enough that it was used in the proof you cite, then it may be that both of you are sloppy.
• I guess I'll go with (3.) then. But I don't understand how you can know that A and B are square? Can't they be any size as long as matrix multiplication is defined? Feb 6, 2015 at 8:56
• @PandaDeTapas That is true. I suggest you re-read the statements to make sure they did not say "let $A$ and $B$ be square matrices" somewhere. If they did not, then... they were sloppy. Of course, if $A=[1,0]$ and $B=[0,1]^T$, then $AB=I$ and $BA\neq I$.
– 5xum
Feb 6, 2015 at 9:00
• No the only words they used was "show that" and then the equivalence expression. No information what so ever about the matrices in question. Feb 6, 2015 at 9:05
• @PandaDeTapas Well, books sometimes make implicit assumptions that are, as mathematitians say, "obvious". I advise you to ask some of your classmates, or even your professor, about this question. If, for example, a lot of them say "yes, of course $A$ and $B$ are square", then it may be that you missed something.
– 5xum
Feb 6, 2015 at 9:32

Assume for exemple $AB=I$. Then $A$ and $B$ represent two linear maps $f$ and $g : \mathbb{R}^n \to \mathbb{R}^n$, $n$ is the size of the matrix. Then you have $f\circ g= id$, which implies easily that $f$ is surjective and $g$ is injective (1). But in finite dimensional linear algebra, those facts are equivalent $f$ and $g$ to be bijective (2), and you are done with the book's proof...

(1) and (2) are exercises, if you have a problem with you can ask, but it's probably solved somewhere in your book.

We assume that $$AB = I$$ and we want to prove that $$BA$$ will be $$I$$ as well. I'll prove by contradiction.

Let's assume that $$AB = I$$, but $$BA = C \neq I$$(a $$C$$ which is not the identity matrix). Then left-multiplying both sides by $$A$$ it gives $$A(BA) = A(C)\,\,\, != A(I)$$. Also since matrix multiplication is associative we can say $$(AB)A = AC \,\,\, != A$$. Now, according to our initial assumption, $$AB = I$$, so we can say $$IA = AC \,\,\, != A$$, but it is $$A\,\,\, != A$$ which is a contradiction, so the $$C$$ must be $$I$$.

• using the cancellation law, which maybe also needs to be proved at this stage. Feb 26, 2021 at 17:14
• You can't multiply an inequality by a matrix. $C\neq I$ does not imply $AC\neq AI$; for example, $0C=0I$. Jan 31, 2023 at 18:21