How to show $(x\cdot y, x\cdot z) = ||x||^2 (y,z) \quad \forall x,y,z \in \mathbb{H}$

In the book of Linear Algebra by Werner Greub, at page 209, it is given that

$$(x\cdot y, x\cdot z) = ||x||^2 (y,z) \quad \forall x,y,z \in \mathbb{H}$$ , which are easily verified using (7.36) and (7.38), where

$$(x\times y, z) = \Delta(x,y,z) \quad \text{7.36} \\ (x_1 \times x_2, y_1\times y_2) = (x_1, y_1)(x_2,y_2) - (x_1, y_2)(x_2)(y_1)\quad x_i, y_j \in E, \quad (7.38)$$ where $E$ is a 3-dimensional oriented Euclidean inner product space.

However, I'm having difficulty in showing this result. I mean the only way that I can think of is decomposing $x,y\in \mathbb{H}$ as $$a = \lambda_a e + a_0, \quad a_0 \in \{e\}^\perp \subseteq \mathbb{H}.$$ However, with this the expression becomes really long and I can't see anything, and also note that the author says "easily verified", so my question is how can we prove this result ?

Note: The definition of the product between quaternions is given as $$e\cdot x = x\cdot e = x \quad x\in \mathbb{H}$$ $$x\cdot y = -(x,y)e + x\times y\quad x,y\in \{e\}^\perp \subseteq \mathbb{H}$$

The quaternion norm is multiplicative:

$$\begin{array}{ll} |(a+\mathbf{u})(b+\mathbf{v})|^2 & =|(ab-\mathbf{u}\cdot\mathbf{v})+(a\mathbf{v}+b\mathbf{u}+\mathbf{u}\times\mathbf{v})|^2 \\ & =|ab-\mathbf{u}\cdot\mathbf{v}|^2+|a\mathbf{v}+b\mathbf{u}+\mathbf{u}\times\mathbf{v}|^2 \\[6pt] & = \qquad a^2b^2 \,- \, \color{Red}{2ab\,\mathbf{u}\cdot\mathbf{v}} ~+~ \color{Blue}{|\mathbf{u}\cdot\mathbf{v}|^2} \\ & \phantom{=}+a^2|\mathbf{v}|^2+\color{Red}{2ab\,\mathbf{u}\cdot\mathbf{v}}+b^2|\mathbf{u}|^2+\color{Blue}{|\mathbf{u}\times\mathbf{v}|^2} \\[6pt] & =a^2b^2+a^2|\mathbf{v}|^2+b^2|\mathbf{u}|^2+|\mathbf{u}|^2|\mathbf{v}|^2 \\ & = (a^2+|\mathbf{u}|^2)(b^2+|\mathbf{v}|^2) \\ & =|a+\mathbf{u}|^2|b+\mathbf{v}|^2. \end{array}$$

Facts used:

• Definition of the norm, $|a+\mathbf{u}|^2=a^2+|\mathbf{u}|^2$.
• Fact $|\mathbf{x}+\mathbf{y}|^2=|\mathbf{x}|^2+|\mathbf{y}|^2$ for orthogonal vectors $\mathbf{x}$ and $\mathbf{y}$.
• expand $(\mathbf{x}+\mathbf{y})\cdot(\mathbf{x}+\mathbf{y})$ or use Pythagorean theorem
• Vector identity $|\mathbf{u}\cdot\mathbf{v}|^2+|\mathbf{u}\times\mathbf{v}|^2=|\mathbf{u}|^2|\mathbf{v}|^2$.
• use $\mathbf{u}\cdot\mathbf{v}=|\mathbf{u}||\mathbf{v}|\cos\theta$ and $|\mathbf{u}\times\mathbf{v}|=|\mathbf{u}||\mathbf{v}|\sin\theta$

The quaternion inner product is $(a+\mathbf{u},b+\mathbf{v})=ab+\mathbf{u}\cdot\mathbf{v}$, which relates to the norm via the identity $|q|^2=(q,q)$. The multiplicativity of the norm can be expressed as

$$(xy,xy)=(x,x)(y,y)$$

This can be polarized. The term polarization refers to the process of switching between quadratic and bilinear gadgets (or generalizations with even more variables). In particular, we can replace $y$ with $y+z$ in the above identity, and then use the bilinearity of the inner product:

$$\begin{array}{ccc} (xy+xz,xy+xz) &= & (x,x)(y+z,y+z) \\ (xy,xy)+2(xy,xz)+(xz,xz) & = & (x,x)\big[(y,y)+2(y,z)+(z,z)\big] \\ (xy,xz) & = & (x,x)(y,z). \end{array}$$

You can even polarize further by replacing $x$ with $w+x$ to get a four-variable identity.

Note polarization is also how we can determine the equivalence between preserving the inner product (i.e. $\langle Ax,Ay\rangle=\langle x,y\rangle$, which preserves lengths and angles as special cases) and just preserving the norm (i.e. $|Ax|=|x|$) as part of the definition of $SO(n)$, the special orthogonal group consisting of $n\times n$ rotation matrices.

Historically, quaternions were discovered (by Hamilton, hence the $\mathbb{H}$) by adjoining (i) a sqrt of $-1$ to $\mathbb{C}$ and (ii) writing out the condition for the norm to be multiplicative. Squaring to $-1$ and multiplicativity were key facts in how complex numbers can model 2D rotation, so the idea was to carry these further into a larger number system to model 3D rotations. It turned out we need the new square root $j$ to anticommute with the first one $i$, and for $ij$ to jut out into a fourth dimension, in order to succeed. Ultimately this leads to quaternions modelling both 3D and 4D rotations.

In this context, the multiplicativity of the norm is not a magical consequence of the definition, it was part of the design of the definition, so the derivation of multiplicativity I gave above wouldn't be necessary. (Although writing everything out with $i$s and $j$s would be, as a trade-off.) One then notices the multiplication table for $i,j,k$ (where $k:=ij$) matches the dot and cross products of vectors $\mathbf{i},\mathbf{j},\mathbf{k}$ in order to get the identity $\mathbf{u}\mathbf{v}=-\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\times\mathbf{v}$ (which becomes invaluable for proving facts about the geometry of quaternions).

The fact that $i,j,k$ have such symmetry - the fact that any oriented real 3D inner product space can be used for a space of imaginary quaternions - is a peculiarity particular to low dimensions. If you try to generalize this construction one way and get Clifford algberas, the next step has zero divisors so certainly the subspace of nonreals doesn't have the same level of symmetry. And if you try to generalize to division algebras, so octonions to be specific, you get a space of imaginary octonions that is somewhat, but not quite as symmetric. (Its symmetry group, $G_2$, has positive codimension in $SO(7)$ - although it acts at least transitively on $2$-frames, so almost.)