Cross-product properties from abstract definition Given two $3$D vectors $\mathbf{u}$ and $\mathbf{v}$ their cross-product $\mathbf{u} \times \mathbf{v}$ can be defined by the property that, for any vector $\mathbf{x}$ one has $\langle \mathbf{x} ; \mathbf{u} \times \mathbf{v} \rangle = {\rm det}(\mathbf{x}, \mathbf{u},\mathbf{v})$.
From this a number of properties of the cross product can be obtained quite easily. It is less obvious that, for instance $|\mathbf{u} \times \mathbf{v}|^2 = |\mathbf{u}|^2 |\mathbf{v}|^2 - \langle \mathbf{u} ; \mathbf{v} \rangle ^2$, from which the norm of the cross-product can be deduced.
Is it possible to obtain these properties nicely (i.e. without dealing with coordinates), but with elementary linear algebra only (i.e. without the exterior algebra stuff, only properties of determinants and matrix / vector multiplication).
Thanks in advance!  

I believe I've found an elegant proof.

*

*Assume that $ \mathbf{u}\times \mathbf{v} \ne 0$. For $\mathbf{x} = \mathbf{u}\times \mathbf{v}$ we have
\begin{align}
\|\mathbf{u}\times \mathbf{v}\|^4 &= \langle \mathbf{u}\times \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle^2 \\
&=\det(\mathbf{u}\times \mathbf{v},\mathbf{u},\mathbf{v})^2\\
&= \det\left(\begin{bmatrix} \mathbf{u}\times \mathbf{v} \\ \mathbf{u} \\ \mathbf{v}\end{bmatrix}\begin{bmatrix} \mathbf{u}\times \mathbf{v} ,\mathbf{u} ,\mathbf{v}\end{bmatrix}\right)\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & \langle \mathbf{u}\times \mathbf{v}, \mathbf{u}\rangle & \langle \mathbf{u}\times \mathbf{v},\mathbf{v}\rangle \\ \langle \mathbf{u},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{u},\mathbf{u}\rangle & \langle \mathbf{u},\mathbf{v}\rangle \\ \langle \mathbf{v},\mathbf{u}\times \mathbf{v}\rangle & \langle \mathbf{v},\mathbf{u}\rangle & \langle \mathbf{v},\mathbf{v}\rangle\end{vmatrix}\\
&= \begin{vmatrix} \|\mathbf{u}\times \mathbf{v}\|^2 & 0 & 0 \\ 0 & \|\mathbf{u}\|^2 & \langle \mathbf{u},\mathbf{v}\rangle \\ 0 & \langle \mathbf{u},\mathbf{v}\rangle & \|\mathbf{v}\|^2\end{vmatrix}\\
&= \|\mathbf{u}\times \mathbf{v}\|^2(\|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2).
\end{align}
Dividing by $\|\mathbf{u}\times \mathbf{v}\|^2$ gives the result.


*Assume that $ \mathbf{u}\times \mathbf{v} = 0$. Then the $\det(\mathbf{x},\mathbf{u},\mathbf{v}) = 0$ for all vectors $\mathbf{x}\in\Bbb{R}^3$ so $\{\mathbf{u},\mathbf{v}\}$ are linearly dependent. Equality condition in Cauchy-Schwartz gives
$$\|\mathbf{u}\times \mathbf{v}\|^2 = 0 = \|\mathbf{u}\|^2\|\mathbf{v}\|^2 - \langle \mathbf{u},\mathbf{v}\rangle^2.$$
