Proof that quaternion multiplication preserves norm

Question

$\require{cancel}$ I know that quaternion multiplication preserves norm.

$N(pq) = N(p)N(q) $

I also know that there is a simple proof for this.

$\sqrt{pq\overline{pq}} = \sqrt{pq\bar{q}\bar{p}} = \sqrt{p(q\bar{q})\bar{p}} = N(p)N(q)$

I am trying another way to prove this.

As far as I know, one can represent a quaternion number as [s, v], s for real and v for imaginary.
Then quaternion multiplication can be obtained by the formula below.
$[s_1, v_1][s_2,v_2] = [s_1s_2 - v_1\cdot v_2, s_1v_2 + s_2v_1 + v_1 \times v_2]$

Let q be [s, v], then the norm of q can be represented as $N(q)=\sqrt{[s, v][s, -v]}=\sqrt{[s^2-(v)\cdot (-v), 0]}=\sqrt{[s^2 + v\cdot v, 0]} = \sqrt{s^2+v\cdot v}$.

Let $p = s_1 + v_1$ and $q=s_2+v_2$ from now on.

As I want to prove that $N(pq) = N(p)N(q)$. Let's do some dirty math.
$N(p)^2 = s_1^2+v_1\cdot v_1$
$N(q)^2 = s_2^2+v_2\cdot v_2$
$N(p)^2N(q)^2 = s_1^2s_2^2+s_1^2(v_2\cdot v_2) +s_2^2(v_1\cdot v_1) + (v_1\cdot v_1)(v_2\cdot v_2)$

\begin{align} N(pq)^2 & = N([s_1s_2 - v_1\cdot v_2, s_1v_2 + s_2v_1 + v_1 \times v_2])^2 \\ & = (s_1s_2-v_1\cdot v_2)^2 + (s_1v_2 + s_2v_1 + v_1 \times v_2)\cdot (s_1v_2 + s_2v_1 + v_1 \times v_2) \\ &= s_1^2s_2^2 \cancel{-2s_1s_2v_1\cdot v_2}+(v_1\cdot v_2 )^2 \\ & \quad\quad + s_1^2(v_2\cdot v_2) + \cancel{s_1s_2v_1\cdot v_2} + \cancel{s_1v_2\cdot(v_1\times v_2)} \\ & \quad\quad + \cancel{s_1s_2v_1\cdot v_2} + s_2^2(v_1\cdot v_1) + \cancel{s_2v_1\cdot(v_1\times v_2)}\\ & \quad\quad + \cancel{s_1v_2\cdot (v_1\times v_2)} + \cancel{s_2v_1\cdot (v_1\times v_2)} + (v_1\times v_2)\cdot (v_1\times v_2)\\ \end{align}
The cancel occurs due to the repeated term and scalar triple product of duplicated vector.
To summarize, we get $N(pq)^2= s_1^2s_2^2+s_1^2(v_2\cdot v_2)+s_2^2(v_1\cdot v_1)+(v_1\cdot v_1)(v_2\cdot v_2)+(v_1\times v_2)\cdot (v_1\times v_2)$

Everything looks fine except the last term and I have no idea what is wrong.
What am I missing here?

Answer 1

I found where I made a mistake.
Let me refer $|v|^2=v\cdot v$ as $v^2$.
I should have noted that $v_1^2v_2^2 \neq (v_1v_2)^2$.
To summarize, $N(p)^2N(q)^2 = s_1^2s_2^2+s_1^2v_2^2+s_2^2v_1^2+v_1^2v_2^2$
$N(pq)^2=s_1^2s_2^2 + s_1^2v_2^2+s_2^2v_1^2+(v_1v_2)^2+(v_1\times v_2)^2$.
Thus, I have to show that $v_1^2v_2^2 = (v_1v_2)^2 + (v_1\times v_2)^2$ now.
We can show this with magnitude property of dot and cross product.

\begin{align} (v_1v_2)^2 + (v_1\times v_2)^2 & =(|v_1||v_2|cos\theta)^2 + (|v_1||v_2|sin\theta)^2 \\ & = |v_1|^2|v_2|^2 \\ & = v_1^2v_2^2 \end{align}

Answer 2

This answer is just an intuitive geometry explanation.

First, by definition we have

$$N(\lambda p) = \lambda N(p)$$

or, in another notion

$$\lVert \lambda p \rVert = \lambda \lVert p \rVert$$

So, we can normalize every quaternion to a unit quaternion or a versor, i.e. for any two quaternion $p$ and $q$

$$u = \frac{p}{\lVert p \rVert} $$ $$v = \frac{q}{\lVert q \rVert} $$

gives us two versors $u$ and $v$, and we need proof the fact that the multiplication $uv$ is still a versor.

Because versor is a 3d rotation, and the multiplication of $uv$ is just applying rotation $u$ followed by rotation $v$. Since 3d rotations form a group, that is to say $uv$ is also a versor.