I need to show
Let $u$ be a unit quaternion such that $|u| = 1$. Show that $ui, uj, uk$ is a basis for $u^\perp$.
We weren't given a definition for what $u^\perp$ is, but I suppose it's similar to that of the orthogonal complement from linear algebra.
I know that to show something is a basis, we need to show that it spans and is linearly independent. I tried showing linear independence first:
$$aui + buj + cuk = 0$$ $$\implies uai + u bj + uck = 0$$ $$\implies u(ai + bj + ck) = 0$$
But then I got stuck. I don't know if this is valid exactly:
$$\left|u(ai + bj + ck)\right| = |u||ai + bj + ck| = 1|ai + bj + ck| = 0$$
We have $$|ai + bj + ck| = a^2 + b^2 + c^2 = 0 \implies a = b = c = 0$$
It seems right. On the assumption that what I did is valid, how might I show that $ui, uj, uk$ spans $u^\perp$?
I know we can follow an example like this, but I don't know what a quaternion in $u^\perp$ looks like exactly. Does it suffice to show that $u \cdot (a_1ui +a_2uj + a_3uk) = 0$?
If so, we would begin something like this: $$u \cdot (a_1ui +a_2uj + a_3uk) = a_1(u \cdot ui) + a_2(u \cdot uj) + a_3(u \cdot uk) = ?$$
I'm not sure what comes next. Can someone provide a hint?
By definition, $u^\perp$ is the space of all elements orthogonal to $u$.
When you have $u(ai+bj+ck)=0$, you can left-multiply both sides by $u^{-1}$ (which is $\overline{u}$ by the way) to get $ai+bj+ck=0$. Note $i,j,k$ are linearly independent by construction, or you can use the norm idea if you want.
Since the quaternions $\mathbb{H}$ are $4$-dimensional and the span of $u$ is $1$-dimensional, its orthogonal complement $u^\perp$ is $3$-dimensional. So a basis will be any three linearly independent elements.
Every element $u(ai+bj+ck)$ is orthogonal to $u$. The key fact is that $\langle ux,uy\rangle=\langle x,y\rangle$ for all quaternions $x,y$ and unit quaternions $u$, where $\langle -,-\rangle$ is the inner product. This follows from the polarization identity $\langle x,y\rangle=(|x+y|^2-|x|^2-|y|^2)/2$. Since left-multiplying by a unit quaternion preserves norms, it must preserve inner products too. Therefore
$$\langle u,u(ai+bj+ck)\rangle=\langle 1,ai+bj+ck\rangle=0. $$