# Does AB=I imply BA=I?

1. Feb 21, 2012

### asmani

Hi all.

Does there exist a field F such that AB=I doesn't imply BA=I, where A and B are square matrices (both n×n) over the field F?

2. Feb 21, 2012

### tiny-tim

hi asmani!

hint: BAB ?

3. Feb 21, 2012

### morphism

AB=I always implies BA=I regardless of the field. This is because the rank-nullity theorem holds over any field, so an injective linear map (between vector spaces of the same dimension n) is necessarily surjective (and vice-verse).

In fact AB=I will imply BA=I even over a commutative ring. Probably the quickest way to see this is to use the characterization "An nxn matrix with entries in a commutative ring R is invertible (i.e. has a left and right inverse) iff detA is a unit in R". Note that AB=I implies that detA and detB are units.

P.S. Here I'm addressing the problem of showing that "left invertible" implies "right invertible". Once you have this, you can use tiny-tim's hint to show that the left inverse is equal to the right inverse.

Last edited: Feb 21, 2012
4. Feb 22, 2012

### vish_maths

AB=I does not always imply BA=I . There are somethings called left inverses and right inverses. You can look it up when you see the least square projections of a vector which does not lie on a column space S onto S.

Then if there is a left inverse C , then, though , CA = I but AC ≠ I .
A similar story goes for the right inverse.

Good Luck

5. Feb 22, 2012

### micromass

We are talking about square matrices. In that case, AB=I does certainly imply BA=I.

6. Feb 22, 2012

### vish_maths

oops :) yep . Sorry , i was talking in general terms :) thanks for quoting !

7. Feb 22, 2012

### D H

Staff Emeritus
Not always. There are certain mathematical characteristics that are required to be able to say that. I would have agreed completely had you instead said "We are talking about square matrices over a field."

This is a very nice hint. What mathematical characteristics let tiny-tim get away with writing BAB without using any parentheses? What characteristics do you get "for free" just by saying the word "field"?

8. Feb 24, 2012

### asmani

Thanks a lot for the replies.
Actually the original problem was to show that "left invertible" implies "right invertible" and vice versa.
I can't see how to work out this yet. I know that we can write B=BI and since AB=I, we get to B=B(AB) which is the same as B=(BA)B (commutativity of multiplication) and then we have B(I-BA)=0. Does it help?

9. Feb 24, 2012

### Bacle2

Careful! If you had commutativity _matrix-wise_ (you do have it entry/element-wise),
AB=I would automatically imply BA=I.

10. Feb 24, 2012

### mathwonk

this is not a trivial result. it is equivalent to showing that any set of n independent vectors in k^n must span it.

11. Feb 25, 2012

### asmani

Sorry, I meant associativity, which leads to B(AB)=(BA)B.
Thanks.

12. Feb 25, 2012

### asmani

Here is the proof I found:

AB=I implies det(A)det(B)=1, which means that det(A)≠0. It's a known fact that A is invertible iff det(A)≠0. Thus A is invertible and has both right and left inverses and they are equal.

Is there something wrong?

13. Feb 25, 2012

### morphism

No, there is nothing wrong with this. In fact I already gave you this exact same proof in my post above.

14. Feb 25, 2012

### asmani

You're right. I didn't notice that, because I wasn't familiar with rings.
Thanks.