
#1
Feb2112, 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 n×n) over the field F? Thanks in advance. 



#3
Feb2112, 02:24 PM

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P: 2,020

AB=I always implies BA=I regardless of the field. This is because the ranknullity theorem holds over any field, so an injective linear map (between vector spaces of the same dimension n) is necessarily surjective (and viceverse).
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 tinytim's hint to show that the left inverse is equal to the right inverse. 



#4
Feb2212, 05:15 AM

P: 47

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 



#5
Feb2212, 06:41 AM

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P: 16,562





#6
Feb2212, 06:58 AM

P: 47





#7
Feb2212, 07:58 AM

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P: 14,436





#8
Feb2412, 12:30 PM

P: 84

Thanks a lot for the replies.




#9
Feb2412, 05:09 PM

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P: 1,168

AB=I would automatically imply BA=I. 



#10
Feb2412, 10:12 PM

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P: 9,421

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
Feb2512, 12:27 AM

P: 84

Thanks. 



#12
Feb2512, 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? 



#13
Feb2512, 04:18 AM

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#14
Feb2512, 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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