Andrei from Romania sent in a good
solution to this problem.
(1)(a) We have to show that $qq^{-1} = 1$: $$q^{-1} = ({1\over
\sqrt 2} + {1\over \sqrt 2}{\bf i}) ({1\over \sqrt 2} - {1\over
\sqrt 2}{\bf i}) = {1\over 2}(1 - {\bf i}^2) = 1$$ (b)Take $x =
ti$ to be any point on the x-axis. Then $qx = ({1\over \sqrt 2} +
{1\over \sqrt 2}{\bf i})t{\bf i} = {-1\over \sqrt 2} + {1\over
\sqrt 2}{\bf i})t = xq.$
We have shown that $qx = xq$ and so $qxq^{-1} = x$.
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Let $F(v) = qvq^{-1}$ be a mapping of points $v$ in
$R^3$ to images in $R^3$ where the 4-dimensional quaternion
$q$ acts as an operator. We have proved that this mapping
fixes every point on the x axis. |
(c)What effect does this mapping have on other points in $R^3$?
$$F({\bf j})= qjq^{-1} = ({1\over \sqrt 2} + {1\over \sqrt
2}{\bf i})({\bf j}) ({1\over \sqrt 2} - {1\over \sqrt 2}{\bf
i})$$ $$ = {1\over 2}(1+{\bf i})({\bf j})((1 - {\bf i})$$ $$ =
{1\over 2}({\bf j} + {\bf k})(1 - {\bf i})$$ $$ = {1\over
2}({\bf j} + {\bf k} + {\bf k} - {\bf j})$$ $$= {\bf k}}.$$
Similarly $F{\bf k} = -{\bf j}$. Parts (b) and (c) together
show that the mapping $F(v) = qvq^{-1}$, where $q = \cos (\pi
/4) + \sin (\pi /4) {\bf i}$, gives a rotation of $\pi /2$
about the x axis.
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(2) In this section
we consider the mapping
G(v) = qvq-1 of
R3 to R3 where
the quaternion q = cosq+ sinqk
is an operator.
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(a)
(cos q+ sinqk)(cosq- sinqk) = cos2 q+ sin2 q = 1 so these two quaternions
are multiplicative inverses.
(b) We have qk = -sinq+ cosqk = kq and hence
qkq-1 = k.
So the mapping G(v) = qvq-1 fixes the z-axis.
(c) What effect does the mapping G have on vectors in R3?
We consider the vector
v = r(cosfi + sinfj).
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(cosq+ sinqk)r(cosfi + sinfj) (cosq- sinqk) |
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r((cosqcosf- sinqsinf)i +(cosqsinf+ sinqcosf)j) cosqi -sinqk) |
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r(cos(q+ f) i + sin(q+ f)j) (cosqi -sinqk) |
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r((cos(q+f)cosq- sin(q+ f)sinf)i + (cos(q+f)sinq+ sin(q+ f)cos phi)j) |
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| r(cos(2q+ f)i + sin(2q+ f)j) |
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We have shown
qkq-1 = k so
G(v+tk) = q(v +tk)q-1 = qvq-1 + qtkq-1 = qvq-1 + tk for all t.
We can see that the vector
v in the xy plane is
rotated about the z
axis by an angle 2q
and all points on the
vertical line through it
are also rotated about
the z-axis by an angle
2q. So by the mapping
G(v) = qvq-1 all points
in R3 are rotated by
2q about the z-axis.
Note that, for any rotation
of R3, we can make a
transformation of the
coordinate system so that
the axis of the rotation
is made to coincide with the
z-axis, then perform the
rotation by the given angle
about the z-axis, and
finally transform back to
the original coordinate
system.
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