If we have set of orthogonal vectors (X) can we form a set of orthogonal bivectors from that set?
I am trying to find if there is a way to get 'more information' from an orthogonal matrix by some sort of manipulation (like the exterior product). If I consider uncorrelated variable as orthogonal matrix can independent variable mean something else?
Additionally, may be I can put it this way: If I have a model of the orthogonal set of vectors which have a unknown factor due to noise, can we learn more on the space of bivectors?
Yes, if the set of vectors $X$ consists of mutually orthogonal vectors, then all bivectors formed from products of elements in this set are mutually orthogonal.
How can we demonstrate this? Consider $a, b, c, d \in X$, $b \neq a$ and $c \neq d$. Then $ab$ and $cd$ are bivectors, using the geometric product. Now consider the product $abcd$.
If $a = c$, then $abad = -aabd$ by orthogonality, and we get $-|a|^2bd$. This only produces a scalar if $b = d$ also; otherwise, we get another bivector (the commutator product).
If $a \neq c$ but $b = d$, the logic is the same as above.
If $a \neq c$ and $b \neq d$, then all four vectors are orthogonal, and we get a grade-4 multivector--one with no scalar part.
Thus, $\langle abcd \rangle_0 \neq 0$ only when $a=c$ and $b =d$, and thus when $ab = cd$. This, to me, is just like the notion of orthogonality in vectors, so I think we can use the same word for this.
I used only the elements of the same set to help restrict some of the cases to be considered, but since bivectors form their own vector space, that should be enough to argue that this notion of bivector orthogonality is just like that of vector orthogonality.