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Matrix Proof

  1. Sep 18, 2012 #1
    An n × n matrix is skew-symmetric provided
    A^T = −A. Show that if A is skew-symmetric and
    n is an odd positive integer, then A is not
    invertible.


    When you do this proof, is it necessary to prove that the determinant of A transpose = determinant of -A?
     
  2. jcsd
  3. Sep 18, 2012 #2

    Dick

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    Why would you need to prove that? You are given that A^T=(-A), so it's pretty obvious that det(A^T)=det(-A). For a general matrix, what's the relation between det(A) and det(A^T)??
     
  4. Sep 18, 2012 #3

    gabbagabbahey

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    No, if [itex]A=B[/itex], then clearly [itex]\det(A)=\det(B)[/itex] (if you do any operation on [itex]A[/itex], then the result must be the same as doing that operation on [itex]B[/itex], otherwise the two are not equal.
     
  5. Sep 18, 2012 #4

    The same. I saw someone (http://answers.yahoo.com/question/index?qid=20090317163121AAkzR4p) do it on yahoo answers in a really complicated way and he proved that those 2 are equal so it got me confused. Like you said you're given one matrix EQUAL to another so I don't understand why that user proved it.
     
  6. Sep 18, 2012 #5
    Generally, it takes me a little longer to do a proof than it would for me to do a calculation (application of proof/theory). I'm not sure if this is the same with everyone. Usually I can end up figuring it out, but sometimes I'm not thinking in the right direction which can end up taking a long time.


    Do you guys know of any sites that have problems where I can get some practice with matrix proofs?
     
  7. Sep 18, 2012 #6

    Dick

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    Why don't you start with this one? It's pretty easy. Answer my first question and then tell me what the relation is between det(-A) and det(A).
     
  8. Sep 18, 2012 #7

    I said it was the same in my other post.

    det(-A) = (-1)^n * det(A)
    using the concept where A is an nxn matrix and a is a constant. Det (-aA) = -a^n * det(A)

    This is how I attempted to do this problem:

    det(A^T)=det(-A)=det(A)
    since n is odd det(-A) = -det(A)
    -det(A)=det(A)
    0 is the the only # where the negative of itself is itself.


    That last part is seemingly easy, but it took me a while to think in that direction.
     
  9. Sep 18, 2012 #8

    Dick

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    Ah, ok. That's what "The same" meant. You've got the proof in your head just fine. Your statement has a funny looking det(-A)=det(A) statement. I assume that's a typo. Actually I see what you mean by that. You mean det(A^T)=det(A) (because it's true for any matrix) and det(A^T)=det(-A) (because the matrix is skew). Putting all of that in one line without giving reasons is confusing and bad style.
     
    Last edited: Sep 18, 2012
  10. Sep 18, 2012 #9
    Oh yeah. I think it would've been better if I had it as det(A)=det(A^t)=det(-A) or in separate lines like you said.

    Do you know of any sites where I can get more practice with proofs?
     
  11. Sep 18, 2012 #10

    Dick

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    Not giving reasons for statements is pretty bad. Putting a bunch of equalities in the same line that have different reasons for being true is asking trouble. You are asking whoever is reading the proof to guess what you are thinking. Don't do it. Make it as clear on paper as it is in your head. Sorry, don't know any practice proof sites.
     
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