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Quaternion Polynomial Equation

  1. Dec 9, 2011 #1
    I'm working on the following:

    "Prove that [itex]x^2 - 1=0[/itex]" has infinitely many solutions in the division ring Q of quaternions."

    The Quaternions are presented in my book in the representation as two-by-two square matrices over ℂ. The book gives that for a quaternion

    (sorry for the terrible notation, I wasn't able to figure out how to do a matrix in here and I didn't want to do the tex by hand)
    [itex]\stackrel{a+bi\ \ c+di}{-c+di\ \ a-bi}[/itex]

    has inverse

    [itex]\frac{1}{a^2+b^2+c^2+d^2}\stackrel{a-bi\ \ -c-di}{c-di\ \ a+bi}[/itex].

    Now if [itex]x^2-1=0[/itex], then clearly [itex]x=x^{-1}[/itex]. This means that the individual components must be equal, so a+bi is a real multiple of its conjugate, which requires that it is real, and that multiple is one, or that it is imaginary, and that multiple is -1. Since the multiple is identically positive (sum of squares) b=d=0, and since the multiple must equal 1, [itex]a^2+c^2=1[/itex]. This gives us

    [itex]\stackrel{a\ \ c}{-c\ \ a}=\stackrel{a\ \ -c}{c\ \ a}[/itex],

    so c=-c=0. Therefore, a=1, which seems to imply that 1 (the quaternion unity) is the unique solution.

    What am I missing?
  2. jcsd
  3. Dec 9, 2011 #2

    Ben Niehoff

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    Are you sure there isn't a typo in the question? I would expect [itex]x^2 + 1 = 0[/itex] to have infinitely many solutions. [itex]x^2 - 1 = 0[/itex] should have two solutions.
  4. Dec 9, 2011 #3


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    if q* is the quaternial conjugate, then qq* is real. note that q-1 = q*/|q|, so from q = q-1, we get q = q*/|q|, implying that q is real.

    but x2 -1 = 0 has just two real solutions, 1 and -1.

    if it IS a typo, consider (ai + bj + ck)2, where a2+b2+c2 = 1 (which has infinitely many solutions (a,b,c)).
  5. Dec 9, 2011 #4
    In fact, it wasn't a typo at all. The problem just before it references the polynomial [itex]x^2-1[/itex] over an entirely different field, and I misread. Apologies, and thanks for your help!
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