Similarity of some vectors

Question

Let $i_{0}, j_{0}, k_{0}$ be in $\mathbb{H}$ such that

$i_{0}^{2} = j_{0}^{2 }= k_{0}^{2} = i_{0} j_{0} k_{0} = -1$

where $\mathbb{H}$ is quaternions.

Then there exist a quaternion $h \in \mathbb{H}$ such that $i_{0} = h^{-1} i h, j_{0} = h^{-1} j h,k_{0 }= h^{-1} k h,$ where $i, j, k$ are the standard imaginary quaternions, i.e., $i = (0, 1, 0, 0) , j= (0, 0 , 1, 0) , k = (0, 0, 0, 1).$ How?

Answer 1

Hints.

1. Show that $i_0(i_0+i)=(i_0+i)i$.
2. Hence show that $h^{-1}i_0h=i$ for some quaternion number $h$. (You should consider two possibilities $i_0\ne-i$ and $i_0=-i$.)
3. So, without loss of generality, we may assume that $i_0=i$. The given conditions now imply that $j_0^2=-1$ and $ij_0=-j_0i$. Show that $j_0=\cos(t)j+\sin(t)k$ for some real number $t$.
4. Find a quaternion $h$ such that $h^{-1}ih=i$ and $h^{-1}j_0h=j$. (The first equation $h^{-1}ih=i$ means $h$ commutes with $i$.) Hence $i_0$ and $j_0$ are simultaneously conjugate to $i$ and $j$ respectively.
5. Finally, since $k_0$ is uniquely determined by $i_0$ and $j_0$, the result follows.