Gram determinant

Let $V$ be an inner product space over a field $k$ with $⟨\cdot ,\cdot ⟩$ the inner product on $V$ (note: since $k$ is not restricted to be either $ℝ$ or $ℂ$, the inner product here shall mean a symmetric bilinear form on $V$). Let $x_1,x_2,\mathrm{\dots },x_n$ be arbitrary vectors in $V$. Set $r_{ij}=⟨x_i,x_j⟩$. The Gram determinant of $x_1,x_2,\mathrm{\dots },x_n$ is defined to be the determinant of the symmetric matrix

Let’s denote this determinant by $\mathrm{Gram}[x_1,x_2,\mathrm{\dots },x_n]$.

Properties.

1. $\mathrm{Gram}[x_1,\mathrm{\dots },x_i,\mathrm{\dots },x_j,\mathrm{\dots },x_n]=\mathrm{Gram}[x_1,\mathrm{\dots },x_j,\mathrm{\dots },x_i,\mathrm{\dots },x_n]$. More generally, $\mathrm{Gram}[x_1,\mathrm{\dots },x_n]=\mathrm{Gram}[x_{\sigma (1)},\mathrm{\dots },x_{\sigma (n)}]$, where $\sigma$ is a permutation on $\{1,\mathrm{\dots },n\}$.

2. $\mathrm{Gram}[x_1,\mathrm{\dots },a{x}_i+b{x}_j,\mathrm{\dots },x_j,\mathrm{\dots },x_n]=a^2\mathrm{Gram}[x_1,\mathrm{\dots },x_i,\mathrm{\dots },x_j,\mathrm{\dots },x_n]$, $a,b\in k$.

3. Setting $a=0$ and $b=1$ in Property 2, we get $\mathrm{Gram}[x_1,\mathrm{\dots },x_j,\mathrm{\dots },x_j,\mathrm{\dots },x_n]=0$.

4. Properties 2 and 3 can be generalized as follows: if $x_i$ (in the $i$th term) is replaced by a linear combination $y=r_1{x}_1+\mathrm{\cdots }+r_n{x}_n$, then

   $$\mathrm{Gram}[x_1,\mathrm{\dots },y,\mathrm{\dots },x_n]=r_i^2\mathrm{Gram}[x_1,\mathrm{\dots },x_i,\mathrm{\dots },x_n].$$

5. Suppose $k$ is an ordered field. Then it can be shown that the Gram determinant is at least 0, and at most the product $⟨x_1,x_1⟩\mathrm{\cdots }⟨x_n,x_n⟩$.

6. Suppose that in addition to $k$ being ordered, that every positive element in $k$ is a square, then the Gram determinant is equal to the square of the volume of the (hyper)parallelepiped generated by $x_1,\mathrm{\dots },x_n$. (Recall that an $n$-dimensional parallelepiped is the set of vectors which are linear combinations of the form $r_1{x}_1+\mathrm{\dots }+r_n{x}_n$ where $0\le r_i\le 1$.)

7. It’s now easy to see that in Property 5, the Gram determinant is 0 if the $x_i$’s are linearly dependent, and attains its maximum if the $x_i$’s are pairwise orthogonal (a quick proof: in the above matrix, $r_{ij}=0$ if $i\ne j$), which corresponds exactly to the square of the volume of the hyperparallelepiped spanned by the $x_i$’s.

8. If $e_1,\mathrm{\dots },e_n$ are basis elements of a quadratic space $V$ over an order field whose positive elements are squares, then $V$ is , or , iff $\mathrm{Gram}[e_1,\mathrm{\dots },e_n]=0$.