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Then up = az, with <math>\begin{pmatrix}u & 0 \\ az & u \end{pmatrix} </math> as the homography expressing [[conjugation (group theory)|conjugation]] of a rotation by a translation p.
== Linear map ==
{{Disputed|date=February 2020}}
The map <math>f:\mathbb H\rightarrow \mathbb H</math>
of quaternion algebra is called linear, if following equalities hold
: <math>f(x+y)=f(x)+f(y)</math>
: <math>f(ax)=af(x)</math>
: <math>x,y\in\mathbb H, a\in\mathbb R</math>
where <math>\mathbb R</math> is real field.
Since <math>f</math> is linear map of quaternion algebra,
then, for any <math>a, b\in\mathbb H</math>, the map
: <math>(afb)(x)=af(x)b</math>
is linear map.
If <math>f</math> is identity map (<math>f(x)=x</math>),
then, for any <math>a, b\in\mathbb H</math>,
we identify tensor product <math>a\otimes b</math> and the map
: <math>(a\otimes b)\circ x=axb</math>
For any linear map
<math>f:\mathbb H\rightarrow \mathbb H</math>
there exists a tensor <math>a\in\mathbb H\otimes\mathbb H</math>,
<math>a=\sum_s a_{s0}\otimes a_{s1}</math>,
such that
: <math>f(x)=a\circ x=(\sum_s a_{s0}\otimes a_{s1})\circ x=\sum_s a_{s0}xa_{s1}</math>
So we can identify the linear map <math>f</math>
and the tensor <math>a</math>.
== The derivative for quaternions ==
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