Determinants and Matrix Inverses Proofs

  1. 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. jcsd
  3. quasar987

    quasar987 4,770
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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.
     
  4. The second problem can be done without determinants (using contradiction) as well.
     
  5. 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?
     
  6. 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).

    The product of finitely many invertible matrices is invertible. So if A was invertible, then A^k would also be invertible.
     
    Last edited: Jul 30, 2005
  7. quasar987

    quasar987 4,770
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    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.



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