New answers tagged quaternions
8
If instead of $K[x]$ you mean the free algebra generated by $K$ and $x$, then — as noted in my comment — $p(x)=ix+xi-j$ has no root in $\Bbb{H}$, because $ix+xi$ always lies in the plane spanned by $\{1, i\}$.
On the other hand, if instead of $K[x]$ you mean the subset of that free algebra consisting of expressions of the form $\sum k_i x^i$, we ...
2
This can be evaluated with an invocation of the product rule for quaternion functions:
$$(fg)'(t) = f'(t)g(t)+f(t)g'(t)$$
namely by considering $f(\tau) = -q(\tau)q^{-1}(t)$ and $g(\tau) = vq(t)q^{-1}(\tau)$, deriving at $\tau = t$. This means the result is $-q'(t)q^{-1}(t)v-vq(t)(q^{-1})'(t)$. Note that we have assumed $q^{-1}$ to be differentiable.
A ...
2
is there any way to find the number of degrees a line must be rotated to become parallel to another line?
Given 2 straight lines in $\Bbb R^3$, one can define the angle between them as the angle between their vector directions (or its complement to $\pi$ if bigger than $\pi/2$).
More precisely, if you have two vectors $v_1=(x_1,y_1,z_1)$ and ...
1
First of all, let's note that what you linked to is a quaternion derived rotation matrix of an axis-angle representation. That is, given a vector $\vec{a}$ and an angle $\theta$, you get a rotation matrix $A$ such that $Ap_1=p_2$. The only way quaternions came into play is during the translation between the axis-angle and the rotation matrix.
Therefore the ...
3
Okay, here you go. The four generators of that group map to the following matrices in $\operatorname{GL}_{12}$.
$$\left(\begin{array}{rrrrrrrrrrrr}
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & ...
4
One idea that saves some work, but still might make your eyes cross is to realize that negatives work as expected, so you don't need to test associativity with 1, -1, or any of the negative i,j,k. That leaves only $\{i,j,k\}^3$ which is 27 tests, each requiring 4 very easy multiplications, 108 easy lookups. If you notice that $i \mapsto j \mapsto k$ is an ...
3
Let's start with the ring $H$ of Hurwitz quaternions (quaternions s.t. either all coefficient are integers or all coefficients are half-integers). The ring $H\otimes\mathbb F_p$ is a quaternion algebra over a finite field, i.e. a matrix algebra $Mat_{2\times 2}(\mathbb F_p)$ (with determinant as the norm).
Now the binary tetrahedral group is the group of ...
6
Lie algebra $sl_2(\mathbb H)$ is easier to define: it's the Lie algebra generated by traceless matrices. In non-commutative case the space traceless matrices is not closed under Lie bracket — that's why dim>12. In fact, $sl_2(\mathbb H)$ is the Lie algebra of matrices with pure imaginary trace.
$SL(2,\mathbb R)\cong Spin(2,1)$, $SL(2,\mathbb C)\cong ...
0
If $q_1$ can be written as a real $c$ and a vector $a$ (so that $q_1 = c + a$), then using rschwieb's method, you get
$$q_1' = c + q_2 a q_2^{-1} = q_2 (c + a) q_2^{-1} = q_2 q_1 q_2^{-1}$$
where only normalization of a quaternion has been used to manipulate the expression. So what you end up doing is back-rotating the whole system by $q_2^{-1}$, rotating ...
0
The whole thing can be done with quaternion arithmetic.
The axis of $q_1$ is $a=\frac{q_1-\overline{q_1}}{2}$, and by computing $q_2aq_2^{-1}$ you will be rotating the vector $a$ into a new direction.
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