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Quaternion Polynomial Equation 
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#1
Dec911, 12:07 AM

P: 192

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 twobytwo 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\ \ abi}[/itex] has inverse [itex]\frac{1}{a^2+b^2+c^2+d^2}\stackrel{abi\ \ cdi}{cdi\ \ a+bi}[/itex]. Now if [itex]x^21=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
Dec911, 02:59 AM

Sci Advisor
P: 1,588

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.



#3
Dec911, 03:41 AM

Sci Advisor
P: 906

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 x^{2} 1 = 0 has just two real solutions, 1 and 1. if it IS a typo, consider (ai + bj + ck)^{2}, where a^{2}+b^{2}+c^{2} = 1 (which has infinitely many solutions (a,b,c)). 


#4
Dec911, 11:21 AM

P: 192

Quaternion Polynomial Equation
In fact, it wasn't a typo at all. The problem just before it references the polynomial [itex]x^21[/itex] over an entirely different field, and I misread. Apologies, and thanks for your help!



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