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Disprove that ABBA = I 
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#1
Jan612, 05:21 AM

P: 74

The task is to prove that for no two matrices A and B, A*B  B*A = I, where I is the identity matrix.
I tried multiplying by the inverses of A or B, but that doesn't seem to lead to a more manageable form. The only way I see this could be done is by writing down all n*n (assuming n by n matrices) linear equations. It's easy to do when n = 2, but the same contradiction may not be as obvious for higher n. I hope there is a more intelligent way to go about this. 


#2
Jan612, 06:11 AM

P: 2,179

What do you know about determinants?



#3
Jan612, 06:23 AM

P: 74

I know that det(AB) = det(BA), but I don't know what are the properties when subtraction is involved. Except for the case when only one line is different.



#4
Jan612, 06:46 AM

Admin
P: 23,600

Disprove that ABBA = I



#5
Jan612, 07:05 AM

P: 74

What I mean is that I don't know what is det(ABBA) even if I do know det(AB) and det(BA).
I'm looking at Sylvester's determinant theorem which looks related, but I still don't see a solution. Now I need to prove that for no M, det(M+I) = det(M) 


#6
Jan612, 08:54 AM

Sci Advisor
HW Helper
Thanks
P: 25,228




#7
Jan612, 09:30 AM

P: 74

So tr(ABBA) = 0 ? Great. Thanks.



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