Interpreting the determinant of matrices of dot products In the Euclidean space $\mathbb{R}^n$ consider two (ordered) sets of vectors $a_1 \ldots a_k$ and $b_1 \ldots b_k$ with $k \le n$.

Question

*

*What is the geometrical interpretation of $\det(a_i \cdot b_j)?$

*Is it true that $\det(a_i\cdot b_j)=\det(a'_p\cdot b'_q)$ if $a_1\wedge \ldots \wedge a_k=a'_1\wedge \ldots \wedge a'_k$ and $b_1\wedge \ldots \wedge b_k=b'_1\wedge \ldots \wedge b'_k?$

Since $\det(a_i\cdot a_j)$ equals the squared $k$-volume spanned by $a_1\ldots a_k$, I guess that $\det(a_i\cdot b_j)$ may be interpreted as the $k$-volume spanned by some kind of projection of $b_1\wedge \ldots \wedge b_k$ (thought of as an oriented $k$-parallelogram) onto $a_1\wedge \ldots \wedge a_k$. For the same reason I would answer affirmatively to the second question.
 A: *

*Given a Euclidean  structure on $\mathbb R^n$, you deduce a canonical euclidean structure on each  $\Lambda ^k\mathbb R^n$.
On pairs of decomposable elements    $a_1\wedge \ldots \wedge a_k, b_1\wedge \ldots \wedge b_k\in\Lambda ^k\mathbb R^n$  it is given by the formula
$$      (a_1\wedge \ldots \wedge a_k\mid b_1\wedge \ldots \wedge b_k)= \det(a_i\cdot b_j)                                                     $$
So your determinant is the scalar product of two vectors, but in a new vector space.
In particular you have the pleasant interpretation that the volume of the parallelipiped spanned by the vectors $a_1, \ldots ,a_k\in V$ is the length of the vector $a_1\wedge \ldots \wedge a_k\in \Lambda ^k\mathbb R^n$.  

*The above interpretation makes it now obvious that given    $$\omega =a_1\wedge \ldots \wedge a_k,  \:\omega '=a'_1\wedge \ldots \wedge a'_k, \eta=b_1\wedge \ldots \wedge b_k, \eta'=b'_1\wedge \ldots \wedge b'_k$$ the equalities  $\omega=\omega'$ and $\eta=\eta'$ imply that   $$(\omega\mid\eta)=\det(a_i\cdot b_j)=(\omega'\mid\eta')=\det(a'_p\cdot b'_q)$$  
