Firstly, you've mentioned that quaternions can be written as matricies. How?

Please could you wrtite matrices going across in rows, not down in cloumns. eg top left, top middle, top right, bottom left, bottom middle, bottom right. Eg again (1,0,0 0,1,0 0,0,1) for the 3 by 3 identity matrix

Secondly, does this make any operation easier/harder? (How do you talk about the sine of a matrix? Is multiplication easier?)

Thirdly, can one just write them as 4D vectors?

Fourthly, is there any equation with quaternions which has none trivial complex/real solutions. (eg q^5 - 2q + 2 =0)

Advert: www.durham.ac.uk/thomas.willis ... Please visit! :-)

-- Tom Willis (thomas.willis@durham.ac.uk), November 06, 2002

Answers

Hello Tom:

Here is the real 4x4 matrix representation of a quaternion:

|t -x -y -z| |x t -z y| |y z t -x| |z -y x t|

I always use this representation in symbolic math packages such as Mathematica. The reason is that one you start tossing in a complex number, such software thinks there is only one type of imaginary unit vector, i, and not three, i, j, and k.

For programming, it is not as efficient, because everything is done sixteen times. I read a paper about what was the minimal number of additions and multiplications required for the task.

For the sine of a matrix, I would either use a series, or the identity on my algebra tools table (URL at end). That was stolen from complex numbers, changing small i to a 3-vector I.

If you want to write them as 4D vectors that only gives partial information in a sense. A vector is a number that can be added to another vector of the same dimension or multiplied by a scalar. You would have to explicitly state the rules for multiplication. One gets the multiplication rules for free with the matrix representation which is cool.

If I am doing things by hand, I write quaternions as (a, B), where the lower case letter is the scalar, and the capital is the 3-vector. The product is (first - last, innie outie cross) (a^2 -B.B, Ba + aB + BxB) = (a^2 - B^2, 2 a B). This way I never have to remember the i j k rules. And it works in ASCII mode!

Isn't there some fundamental theorem that says any polynomial of the form c^5 - 2c + 2 = 0 will have a solution? I cannot remember what it is called. Oops. Same one applies to quaternions. That's the power of a topological algebraic field.

doug

-- Doug Sweetser (sweetser@theworld.com), November 06, 2002.

Doug alludes to a fundamental theorem that assures the existence of solutions of polynomial equations for real/complex arguments. This is called the Fundamental Theorem of Algebra. It was for a long time just a conjecture. Gauss gave a rigorous proof (maybe several rigorous proofs) about two hundred years ago.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 06, 2004.

Does the fundamental theorum of algebra apply to all fields, or only to those which are are isomorphic to the complex field? Or is it something in between. My module that includes fields stops just short of allowing me to answer this.

-- Tom Willis (thomas.willis@dur.ac.uk), March 06, 2004.

Hello, Thomas!

The Fundamental Theorem Of Algebra does not apply to real fields. The simple counter-example of 0 = 1 + x^2 shows this. If I recollect correctly, Niven proved in the 1940s that the division algebra of quaternions obeys the Fundamental Theorem Of Algebra, though the quaternions may be regarded as not being a field due to noncommutativity of multiplication.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Hello Matthew:

Cool to hear about Niven's result. Let there be no doubt that quaternions are a finite dimensional field. One needs the operations of addition, subtraction, multiplication and division. It is the last one that is tough. The noncommutitativity is a separate issue.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Hi, Doug,

The form of the F.T. of A. that quaternions obey might be called the weak form. It assures the existence of at least one root but does not set a limit on how many distinct roots there may be. The strong form of the F.T. of A., valid for the complex field, would put an upper limit on the distinct roots.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Hello Matthew:

This raises the question of what sort of expression would limit quaternion solutions? It would have to be an expression that contained a curl, because that is antisymmetric, the first step outside the world where everything always commutes as is the case for real and complex numbers.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Hello,Doug,

Limitations can be set on the roots of quaternionic polynomial equations without resorting to field operations such as curl, divergence, gradient, etc.

Finitude can be gotten by requiring sets of simultaneous equations of quaternionic polynomials. This just expresses an algebraic condition.

Best regards,

Matthew

-- Matthew McCann PE (mmccann@franciscan.edu), March 07, 2004.

Good point. Anyway I should have said "cross product," my bad.

doug

-- Douglas Sweetser (sweetser@alum.mit.edu), March 07, 2004.

Consider matices of the form [[a -b] [b a]]. This subset of M_2x2 is isomorphic to the complex plane. So for a polynomial of degree n, there exist at least n solutions in M_2x2. My questions are: Do other solutions exist? Are these solutions equivalent to quaternion solutions? And is there a general formula for finding these roots? Also do solutions exist in M_mxm?

-- Dan McNeill (dmcneill53@yahoo.com), June 28, 2004.

If 2x2 real matrices (denote MR2) are isomorphic to the complex numbers (denote C), then then FTA means that a polynomial of degree n in M2 has exactly n solutions.

Now, I am almost certain that 2x2 matricies with entries that are themselves 2x2 real matricies (denote M(MR2)) is isomorphic to 4x4 real matricies (denote MR4). The isomorphism is the obvious one - remove the brackets round the 2x2 bits.

Quaretnions (denote Q) are isomorphic to a sub-ring of 4x4 real matricies. (I think they form a a skew field? Somehow sub-skew-field seems contrived... but you get the idea).

Writing MC2 for 2x2 complex matricies, "<" for "is a sub-thingy of", and "=" for "is isommorphic to, we have: Q < MR4 = M(MR2) = MC2 (Quaternions are a subset of 4x4 real matricies, which are isomorphic to 2x2 marticies with entires that are 2x2 real matricies, whihc is isomorphic to 2x2 complex matirices).

So quarternions are isomorphic to a sub-set of 2x2 complex matricies... which I think has been stated elsewhere. (Exercise for someone... given a quaternion t + xi + yj + zk, what is its 2x2 complex matrix equivilant?)

There are two questions that seem to be unaswered: give a quaternion polynomial of degree n, what is the upper bound on the number of sloutions it has? The lower bound is 1. As quarternions form a ring of infinate order, surely we get something about the max number of solutions from that? The other question that has come it is: given a polynomial of degree d in nxm matricies with real entries, how many solutions does it have?

-- Tom Willis (thomas.willis@dur.ac.uk), June 29, 2004.