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Coninvolution of a Matrix

  1. Nov 28, 2012 #1
    It's very difficult for me to find any simple literature to explain this idea.

    J[itex]\in[/itex]Mn(ℂ) is a coninvolutory (or a "coninvolution") if A-1=[itex]\overline{A}[/itex]
    I'm looking to prove this lemma:

    Let A be an element of Mn(ℂ) and A is nonsingular, then [itex]\bar{A}[/itex]-1A is coninvolutory.

    I see that the identity matrix is a coninvolution. Does anyone have another example?
     
  2. jcsd
  3. Nov 28, 2012 #2
    It is unclear what you are asking for. Are you asking
    1) How to prove your lemma,
    2) for examples of coinvolutory matrices, or
    3) for general litterature on coinvolutory matrices?

    The lemma is fairly straightforward to prove by calculating [itex]\left(\left(\overline{A}\right)^{-1}A\right)^{-1}[/itex] and showing that it is equal to [itex]\overline{\left(\overline{A}\right)^{-1}A}[/itex] using identities like
    [tex](AB)^{-1}=B^{-1}A^{-1} \qquad \overline{AB} = \overline{A}\,\overline{B}[/tex]
    If you are having trouble proving it, then tell us where you get stuck.

    If you need examples of coinvolutory matrices, then just pick some nonsingular matrix A and calculate [itex]\left(\overline{A}\right)^{-1}A[/itex] as the lemma suggests. Alternatively if you write A = B + i C for real matrices B and C, then you can show that A is coinvolutory precisely if A^2 + B^2 = I and AB=BA. In particular if A is a real matrix, then it is coinvolutory if and only if it is involutory (i.e. iff A^2=I).

    If you want to read more about coinvolutory matrices, then I can't help you as I have never heard of the term and a quick google search does not reveal much.
     
  4. Nov 28, 2012 #3
    I was asking for help on all three and your post really helped a lot! I'll try to be more clear in the the near future. Thank so much!
     
  5. Nov 29, 2012 #4
    Do you mean B^2 +C^2 = I and BC=CB?
     
  6. Nov 29, 2012 #5
    Yes.
     
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