AB=I vs. BA=I: Investigating Matrix Equality in Field F

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In summary, the conversation discusses whether AB=I always implies BA=I for square matrices over a field. It is proven that AB=I indeed implies BA=I over any field, as long as the matrices have the same dimension n. The discussion also touches on the concepts of left and right inverses, as well as the role of commutativity in this problem. Finally, a proof is provided using the determinant and the fact that A is invertible if and only if det(A)≠0.
  • #1
asmani
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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?

Thanks in advance.
 
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  • #2
hi asmani! :smile:

hint: BAB ? :wink:
 
  • #3
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.
 
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  • #4
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
vish_maths said:
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.
 
  • #6
micromass said:
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 !
 
  • #7
micromass said:
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."
tiny-tim said:
hint: BAB ? :wink:
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
Thanks a lot for the replies.
morphism said:
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.
tiny-tim said:
hint: BAB ? :wink:
D H said:
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?
 
  • #9
asmani said:
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.
 
  • #10
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
Bacle2 said:
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.
 
  • #12
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
asmani said:
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.
 
  • #14
You're right. I didn't notice that, because I wasn't familiar with rings.
Thanks.
 

1. What is the meaning of AB=I and BA=I?

AB=I and BA=I are mathematical equations used to represent the multiplication of two matrices, A and B, that results in the identity matrix, denoted by I. This means that when matrix A is multiplied by matrix B, or vice versa, the resulting matrix is the identity matrix.

2. Is AB=I the same as BA=I?

No, AB=I and BA=I are not the same. While both equations result in the identity matrix, the order of multiplication is different. AB=I means that A is multiplied by B, while BA=I means that B is multiplied by A.

3. Can AB=I and BA=I be true for any matrices A and B?

No, for AB=I and BA=I to be true, the matrices A and B must have certain properties. They must be square matrices, meaning they have the same number of rows and columns, and they must be inverses of each other. This means that when A is multiplied by B, the result is the identity matrix, and when B is multiplied by A, the result is also the identity matrix.

4. What does it mean if AB=I and BA=I are true?

If AB=I and BA=I are true, it means that the matrices A and B are inverses of each other. This is a special property in linear algebra and has various applications, such as solving systems of linear equations and finding the inverse of a matrix.

5. Can AB=I and BA=I be true for non-square matrices?

No, AB=I and BA=I can only be true for square matrices. Non-square matrices do not have an inverse, and thus cannot satisfy the conditions for these equations to be true.

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