Maths - Quaternions - open forum

By: nobody ( Nobody/Anonymous )
Quaternion error
2003-03-16 10:26
On this page:

https://www.euclideanspace.com/maths/algebra/realNormedAlgebra/quaternions/index.htm

you say: "In general:

z1 * z2 = conj(z2 * z1)

Where:
Conjugate: q' = conj(q) = a - b i - c j - d k

In other words reversing the order of the operands gives the conjugate of the result that we would get without reversing the operands."

That is not correct. After all, if you let z1 = 1, then the equation reduces to z2 = conj(z2). What is correct is that

conj(z1*z2) = conj(z2) * conj(z1).

-- James W. Walker

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By: martinbaker ( Martin Baker )
RE: Quaternion error
2003-03-16 16:36
James,

Thank you very much for correcting me, I have changed the page to show the equation that you give.

I have added a proof of this, just by brute force multiplying out all the terms, I don't know if there is a more elegant proof?

Cheers,

Martin

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From: "James W. Walker"
To: "Martin.Baker"
Subject: Re: [mjbworld - Open Discussion] Quaternion error
Date: 16 March 2003 21:00

On Sunday, March 16, 2003, at 10:45 AM, Martin.Baker wrote:

> I have added a proof of this, just by brute force multiplying out all
> the
> terms,
> I don't know if there is a more elegant proof?

I can think of a couple of alternate approaches. What's better is a matter of taste.

You could introduce the representation of a quaternion as a pair (s, v) where s is a real scalar and v is a 3D vector. Then conj((s,v)) = (s, -v), and multiplication is given by the formula

(s, v)*(t, w) = (st - v . w, sw +tv + v x w)

where the . is dot product and the x is cross product. With this notation, it's easier to verify the conjugate product equation.

Alternately, you could first observe that conjugation is linear, i.e., if s and t are real and q and r are quaternions, then

conj(sq + tr) = s conj(q) + t conj(r).

Then, assuming you know that quaternion multiplication distributes over addition, it is straightforward to show that the function f(p, q) = conj(p*q) - conj(q) * conj(p) is linear in each variable. Once you observe that f(p, q) = 0 for the 16 combinations of basic quaternions (p and q are 1, i, j, or k), it follows that f(p, q) = 0 for all quaternions.

By the way, it might be a good idea to run your pages through a spelling checker. On this quaternion page, I see the misspellings: dimentional, multiplcations, Normalise (unless this is a British spelling), Infomation, multipication, Matricies, puchased, seperate.