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Determinants and Matrix Inverses Proofs

by KataKoniK
Tags: determinants, inverses, matrix, proofs
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KataKoniK
#1
Jul29-05, 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.
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quasar987
#2
Jul30-05, 12:00 AM
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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.
Muzza
#3
Jul30-05, 03:31 AM
P: 696
The second problem can be done without determinants (using contradiction) as well.

KataKoniK
#4
Jul30-05, 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?
Muzza
#5
Jul30-05, 12:25 PM
P: 696
det(AB) = det(-BA)
detA detB = det(-B) detA
detA detB = (-1)^n detB detA
1 = -(1)^n
You've divided by det(A)det(B), which you can't do, since det(A)det(B) might be 0 (and it is 0).

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

What to do next? Muzza, how would you do this using contradiction?
The product of finitely many invertible matrices is invertible. So if A was invertible, then A^k would also be invertible.
quasar987
#6
Jul30-05, 01:00 PM
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Quote Quote by KataKoniK
det(AB) = det(-BA)
detA detB = det(-B) detA
detA detB = (-1)^n detB detA
1 = -(1)^n
Instead of jumping to the conclusion 1 = -1, what you should have done is ask yourself, when is the equation detA detB = -detB detA satisfied? Only when detA detB = 0 iff detA = 0 or detB = 0 (or both).

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.



Quote Quote by KataKoniK

det(A^k)
= (detA)^k

What to do next?
You're missing the main idea of the proof; the tactic, if you will. The tactic is that you start with an matrix equation that you know to be true. Then taking the determinant of both sides is still a true equation. From there, simplify using determinant properties until you find the desired result. So what you should have done here is

A^k = 0
det(A^k) = det0
(detA)^k = 0 iff detA = 0.
KataKoniK
#7
Jul30-05, 06:56 PM
P: 168
Thanks a bunch guys. It's all clear now.


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