# How do quaternions represent rotations?

I wonder how $qvq^{-1}$ gives the rotated vector of $v$. Is there any easy-to-understand proof for it?

I was on Wikipedia, but I could not understand the proof there because of the conversions.

Why is $uv-vu$ the same as $2(u \times v)$, and why is $uvu$ the same as $v(uu)-2(uv)u$?

-

A (real) quaternion is a "number" of the form $q=a+bi+cj+dk$ where the coefficients $a$, $b$, $c$ and $d$ are real numbers and $i^2=j^2=-1$, $ij=k$, $jk=i$, $ki=j$, $ji=-k$ and so on.

The conjugate quaternion is $\overline{q}=a-bi-cj-dk$, and the (reduced) quaternionic norm of $q$ is the real number $q\overline{q}$.

Pure quaternions, i.e. those for which $a=0$ or equivalently $\overline{q}=-q$ form a 3-dimensional real space, an obvious basis being $\{i,j,k\}$. The quaternionic norm restricted to the space of pure quaternions turns out to be simply the euclidean norm.

The transformation $u\mapsto quq^{-1}$ preserves the space of pure quaternions and preserves the norm, thus can be read as an orthogonal tranformation of ${\Bbb R}^3$ which moreover preserves orientation.

One concludes recalling that an orientation-preserving orthogonal tranformation of ${\Bbb R}^3$ is rotation around some axis.

-
What does the adjective "reduced" signify in "(reduced) quaternionic norm"? – Jason DeVito Feb 18 '11 at 13:50
Well, if you have a finite algebra $A$ over a field $F$, the norm of an element $a\in A$ is the determinant of the multiplication by $a$ thought of as an $F$-linear map $A\rightarrow A$. In the quaternion case, reduced means that instead of taking this as the norm, you take its square root. Since the quaternions are 4-dimensional over $\Bbb R$, the reduced norm defines a quadratic form, which is what one would expect from an euclidean norm. – Andrea Mori Feb 18 '11 at 15:56
But I am still don't understand the 'mechanic' under the quaternion rotation ( why qv(q^-1) gives rotated vector ). And I am still curios how the mentioned conversions (uv−vu ->2 (u×v) and uvu -> v(uu)−2(uv)u) work. – user7217 Feb 18 '11 at 20:50

A proof is outlined here, although I skipped a few computations you should verify.

-
Not that I get any of it, but "maximally nice?" What is maximally nice? – bobobobo Oct 9 '11 at 21:46
@bobobobo: it was a mathematical figure of speech. I just mean "extremely nice." – Qiaochu Yuan Oct 9 '11 at 21:59