1
$\begingroup$

This is a naive question but I read popular science articles where it is stated that quaternions define vector division, without further explanations

$\endgroup$

closed as unclear what you're asking by Eevee Trainer, user21820, Paul Frost, Xander Henderson, Yanior Weg Apr 19 at 21:11

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Can you share a reference to any of these articles? $\endgroup$ – Rahul Apr 17 at 5:07
  • 1
    $\begingroup$ Because vectors can be viewed as quaternions, and we can divide a quaternion with another, this is true in a sense. BUT 1) the result of the division is in general a quaternion rather than a vector. 2) We cannot make sense of quaternionic fractions like $\dfrac{q_1}{q_2}$. We can only use $q_2^{-1}q_2$ or $q_1q_2^{-1}$. In other words, the outcome depends on whether you divide $q_1$ by $q_2$ from the left side or from the right, so we need to specify which way we want to do it. $\endgroup$ – Jyrki Lahtonen Apr 19 at 19:19
2
$\begingroup$

"Yes", but in a very qualified sense. (That is, what @SZN wrote is true, but I'd put a different emphasis on things.)

First, we could ask about "division" of two-dimensional (real) vectors. The complex numbers (which form a field) give a way to make sense of this.

In three dimensions? It turns out that there is no real-three-dimensional field, so, "no".

In four dimensions? Skipping over the three dimensional case was part of Hamilton's inspiration: yes, the Hamiltonian quaternions for a "non-commutative field", also called "skew field" or "division ring" or "division algebra".

But there are no more (real) division algebras than these (by Frobenius' theorem or others). There are further complicated objects (e.g., "octonians"), but these (by the theorem) cannot quite be division algebras.

So, _in_general_, no, there's no reasonable definition of division for vectors, but in some special cases there is.

$\endgroup$
1
$\begingroup$

No. Quaternions can be regarded a specific example of vectors, since the set of quaternions forms a vector space that is isomorphic to $\mathbb{R}^4$ under quaternion addition and scalar multiplication, but a "vector", broadly defined, is any object that belongs to some vector space. The general axioms for an abstract vector space do not allow division and so some vector spaces (e.g. the real or complex numbers) might have a natural division operator that can be defined, while other vector spaces (e.g. the set of 3x5 real matrices) may not. If you are concerned with three-dimensional vectors as used in mechanics, these do not have a natural division operation.

The most general construction that allows division (to my knowledge--but I'm not an algebraist) is a division ring. The most commonly used construction that allows division is a field. The quaternions themselves are often worked with as a division algebra over the field of real numbers and are important because there are only a finite number of such algebras (up to isomorphism) with several nice properties (see the Frobenius theorem). You might also be interested in looking at the Cayley-Dickson construction and Clifford Algebras.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.