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Let $v_1,v_2$ be vectors in $\mathbb{R}^4$. Let $M$ be the $2\times 4$ matrix with rows $v_1,v_2$ in this order. The Gram determinant of $M$ is defined as the determinant of the $2\times 2$ matrix $MM^*$.

For each subset $\sigma$ of $S=\{1,2,3,4\}$ of two elements, let $x_{\sigma}$ be the determinant of the submatrix of $M$ by deleting the $i$-th columns for $i\in \sigma$.

I can prove (and it should be known) that the Gram determinant of $M$ is equal to the sum of $x_{\sigma}^2$ over all subsets $\sigma$ of $S$ of two elements.

My question is: What is the geometric meaning of Gram determinant in this case? It is the square of the length of a vector in $\mathbb{R}^6$, since there are six such subsets $\sigma$. But what is the connection between this vector and $v_1,v_2$?

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up vote 2 down vote accepted

Suppose you have a $m\times n$ matrix $A$ given row-wise by $$A = \begin{pmatrix}\mathbf{a_1} \\ \vdots \\ \mathbf{a_m}\end{pmatrix}$$ then the determinant of the matrix $AA^\mathrm{T}$ gives the square of the hypervolume of the $m$-dimensional parallelotope spanned by the row vectors $\mathbf{a_i}$, embedded in $n$-dimensional space. This is a rather standard geometric interpretation of the determinant in fact. In particular, this gives the geometrical interpretation of the Gram-Determinant as follows:

If the set of vectors comprising the rows of $A$ is linearly independent, then the parallelotope is non-degenerate and so the determinant is non-zero (because the parallelotope has non-zero hypervolume). If on the other hand, the vectors are linearly dependent, then the parallelotope will be degenerate (for example, a cube collapsed into a square) and so it has zero hypervolume.

As for the subset sum you mentioned, are you referring to the Cauchy-Binet formula? If so, you will probably be interested in the geometric considerations resulting from that formula as well. See the above link.

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@Berci Yes, my mistake. I shall edit that. – EuYu Oct 12 '12 at 23:53

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