
#1
Jul2905, 08:07 PM

P: 168

Can anyone help me start this out? I got absolutely no clue.
Q: If A and B are n x n matrices, AB = BA, and n is odd, show that either A or B has no inverse. I know that we have to show that either det A is 0 or det B is 0, but I have no clue how to show it with the given information. Q: If A^k = 0 for some k >= 1, show that A is not invertible. Again, same problem. Any help would be great, thanks. 



#2
Jul3005, 12:00 AM

Sci Advisor
HW Helper
PF Gold
P: 4,768

1: You're right. You gotta show that either detA =0 or detB=0. Take the determinant of both sides of the equation and "simplify" using the properties of the determinant to obtain the desired conclusion.
2: This one's even easier. Use the same trick. 



#3
Jul3005, 03:31 AM

P: 696

The second problem can be done without determinants (using contradiction) as well.




#4
Jul3005, 10:02 AM

P: 168

Determinants and Matrix Inverses Proofs
Thanks for the responses, so is this correc then? I tried it, but I am stuck.
Q: If A and B are n x n matrices, AB = BA, and n is odd, show that either A or B has no inverse. det(AB) = det(BA) detA detB = det(B) detA detA detB = (1)^n detB detA 1 = (1)^n Is this correct? Because since we know n is odd, that means 1 will never equal 1. Is this proof good enough to say that A or B has no inverse? I seriously do not know how to show that either detA or detB is 0. Q: If A^k = 0 for some k >= 1, show that A is not invertible. det(A^k) = (detA)^k What to do next? Muzza, how would you do this using contradiction? 



#5
Jul3005, 12:25 PM

P: 696

Instead, notice that (1)^n = 1, since n was odd. Thus det(A)det(B) = det(A)det(B). 



#6
Jul3005, 01:00 PM

Sci Advisor
HW Helper
PF Gold
P: 4,768

Instead, what you did is you mechanically continued your arithmetics, and divided both sides by detA detB, assuming neither detA and detB was 0...which of course led to the contradiction 1 = 1. A^k = 0 det(A^k) = det0 (detA)^k = 0 iff detA = 0. 



#7
Jul3005, 06:56 PM

P: 168

Thanks a bunch guys. It's all clear now.



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