Let $E$ be an Euclidian oriented vector space of dimension $3$ and $x,y,u,w \in E$.

How do we prove (without coodinates) $$ \det \begin{pmatrix} \langle x,u \rangle & \langle x,w \rangle \\ \langle y,u \rangle & \langle y,w \rangle \\ \end{pmatrix} = \langle x \times y, u \times w \rangle$$

This equality is claimed in this proof of the double cross product formula.

Related : wiki claims that the Gram determinant is equal to : $Gram(x_1,\dots,x_n) = \| x_1 \wedge \dots \wedge x_n \|^2$

How is defined the norm on the exterior product, and how to prove this formula ?

  • $\begingroup$ It should be $⟨x∧y,u∧w⟩$, $x∧u$ will never produce $⟨x,u⟩$ as part of a general Gramian matrix. $\endgroup$ – LutzL May 22 '14 at 14:05
  • $\begingroup$ Right, there are some typos. $\endgroup$ – user10676 May 22 '14 at 15:45

I'm not really sure if you can prove it or simply take it as definition of the scalar product. You can derive it from the definition of a scalar product on the tensor space.

Or just notice that the Gramian determinant $$ \det\begin{pmatrix} \langle x,u \rangle & \langle x,w \rangle \\ \langle y,u \rangle & \langle y,w \rangle \\ \end{pmatrix} $$ is already symmetric in $x∧y$ and $u∧w$. Considering it as a function in $x$ and $y$, $$ \phi(x,y)=\det\begin{pmatrix} \langle x,u \rangle & \langle x,w \rangle \\ \langle y,u \rangle & \langle y,w \rangle \\ \end{pmatrix} $$ one notices that it is bilinear, anti-symmetric, and so (extendable as) a linear form on $\Lambda^2E$. Further, $$ \phi(u,w)=\langle u,u \rangle \langle w,w \rangle - \langle u,w \rangle^2=\|u\|^2\|w-\lambda u\|^2 $$ where $λ=\frac{\langle u,w \rangle}{\|u\|^2}$, tells us that this quadratic form is the square of the area of the parallelogram spanned by $u$ and $v$, which proves its positive definiteness.

Of course, all this reasoning is greatly simplified by using the Binet-Cauchy-formulas for the coordinates of the vectors.


A proof using clifford algebra.

First, write the dot product of two cross products in terms of bivectors.

$$(x \times y) \cdot (u \times w) = (-i)(-i) (x \wedge y) \cdot (u \wedge w)$$

Then, break down the inner product of bivectors into a vector-bivector product and a vector-vector inner product.

$$(x \wedge y) \cdot (u \wedge w) = ([x \wedge y] \cdot u) \cdot w$$

nb. These are actually wedge products I'm using, not cross products.

Then, apply the BAC-CAB rule:

$$[x \wedge y] \cdot u = x (y \cdot u) - y (x \cdot u)$$

The result is

$$([x \wedge y] \cdot u) \cdot w = (x \cdot w)(y \cdot u) - (y \cdot w)(x \cdot u)$$

which is identical to the determinant you were asked to find (within the minus sign that I dropped at the beginning for clarity).

  • $\begingroup$ How to do you prove the BAC-CAB rule? I was first looking to prove BAC-CAB (it is called double crossed product formula in french), but it seemed more natural for me to ask in term of Gram determinant. $\endgroup$ – user10676 May 22 '14 at 15:51
  • $\begingroup$ You can prove the BAC-CAB rule using cyclic permutations of the vectors. I did this in math.stackexchange.com/a/305315/45296 as an example. It's a bit of a pain and not clearly better than breaking down into components and using a basis, in my opinion. $\endgroup$ – Muphrid May 22 '14 at 16:36

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