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Does AB=I imply BA=I?

by asmani
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asmani
#1
Feb21-12, 01:47 PM
P: 84
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 nn) over the field F?

Thanks in advance.
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tiny-tim
#2
Feb21-12, 02:22 PM
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Thanks
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hi asmani!

hint: BAB ?
morphism
#3
Feb21-12, 02:24 PM
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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.

vish_maths
#4
Feb22-12, 05:15 AM
P: 53
Does AB=I imply BA=I?

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
micromass
#5
Feb22-12, 06:41 AM
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Quote Quote by vish_maths View Post
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
We are talking about square matrices. In that case, AB=I does certainly imply BA=I.
vish_maths
#6
Feb22-12, 06:58 AM
P: 53
Quote Quote by micromass View Post
We are talking about square matrices. In that case, AB=I does certainly imply BA=I.
oops :) yep . Sorry , i was talking in general terms :) thanks for quoting !
D H
#7
Feb22-12, 07:58 AM
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Quote Quote by micromass View Post
We are talking about square matrices. In that case, AB=I does certainly imply BA=I.
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."


Quote Quote by tiny-tim View Post
hint: BAB ?
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"?
asmani
#8
Feb24-12, 12:30 PM
P: 84
Thanks a lot for the replies.
Quote Quote by morphism View Post
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.
Actually the original problem was to show that "left invertible" implies "right invertible" and vice versa.
Quote Quote by tiny-tim View Post
hint: BAB ?
Quote Quote by D H View Post
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"?
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?
Bacle2
#9
Feb24-12, 05:09 PM
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Quote Quote by asmani View Post
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?
Careful! If you had commutativity _matrix-wise_ (you do have it entry/element-wise),
AB=I would automatically imply BA=I.
mathwonk
#10
Feb24-12, 10:12 PM
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this is not a trivial result. it is equivalent to showing that any set of n independent vectors in k^n must span it.
asmani
#11
Feb25-12, 12:27 AM
P: 84
Quote Quote by Bacle2 View Post
Careful! If you had commutativity _matrix-wise_ (you do have it entry/element-wise),
AB=I would automatically imply BA=I.
Sorry, I meant associativity, which leads to B(AB)=(BA)B.
Thanks.
asmani
#12
Feb25-12, 12:48 AM
P: 84
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?
morphism
#13
Feb25-12, 04:18 AM
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Quote Quote by asmani View Post
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?
No, there is nothing wrong with this. In fact I already gave you this exact same proof in my post above.
asmani
#14
Feb25-12, 09:54 AM
P: 84
You're right. I didn't notice that, because I wasn't familiar with rings.
Thanks.


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