# Gram determinant

Let $V$ be an inner product space  over a field $k$ with $\langle\cdot,\cdot\rangle$ the inner product  on $V$ (note: since $k$ is not restricted to be either $\mathbb{R}$ or $\mathbb{C}$, the inner product here shall mean a symmetric bilinear form  on $V$). Let $x_{1},x_{2},\ldots,x_{n}$ be arbitrary vectors in $V$. Set $r_{ij}=\langle x_{i},x_{j}\rangle$. The Gram determinant  of $x_{1},x_{2},\ldots,x_{n}$ is defined to be the determinant  of the symmetric matrix  $\begin{pmatrix}r_{11}&\cdots&r_{1n}\\ \vdots&\ddots&\vdots\\ r_{n1}&\cdots&r_{nn}\end{pmatrix}$

Let’s denote this determinant by $\operatorname{Gram}[x_{1},x_{2},\ldots,x_{n}]$.

1. 1.

$\operatorname{Gram}[x_{1},\ldots,x_{i},\ldots,x_{j},\ldots,x_{n}]=% \operatorname{Gram}[x_{1},\ldots,x_{j},\ldots,x_{i},\ldots,x_{n}]$. More generally, $\operatorname{Gram}[x_{1},\ldots,x_{n}]=\operatorname{Gram}[x_{\sigma(1)},% \ldots,x_{\sigma(n)}]$, where $\sigma$ is a permutation  on $\{1,\ldots,n\}$.

2. 2.

$\operatorname{Gram}[x_{1},\ldots,ax_{i}+bx_{j},\ldots,x_{j},\ldots,x_{n}]=a^{2% }\operatorname{Gram}[x_{1},\ldots,x_{i},\ldots,x_{j},\ldots,x_{n}]$, $a,b\in k$.

3. 3.

Setting $a=0$ and $b=1$ in Property 2, we get $\operatorname{Gram}[x_{1},\ldots,x_{j},\ldots,x_{j},\ldots,x_{n}]=0$.

4. 4.

Properties 2 and 3 can be generalized as follows: if $x_{i}$ (in the $i$th term) is replaced by a linear combination  $y=r_{1}x_{1}+\cdots+r_{n}x_{n}$, then

 $\operatorname{Gram}[x_{1},\ldots,y,\ldots,x_{n}]=r_{i}^{2}\operatorname{Gram}[% x_{1},\ldots,x_{i},\ldots,x_{n}].$
5. 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 $\langle x_{1},x_{1}\rangle\cdots\langle x_{n},x_{n}\rangle$.

6. 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 $x_{1},\ldots,x_{n}$. (Recall that an $n$-dimensional parallelepiped is the set of vectors which are linear combinations of the form $r_{1}x_{1}+\ldots+r_{n}x_{n}$ where $0\leq r_{i}\leq 1$.)

7. 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 $i\neq j$), which corresponds exactly to the square of the volume of the hyperparallelepiped spanned by the $x_{i}$’s.

8. 8.

If $e_{1},\ldots,e_{n}$ are basis elements of a quadratic space $V$ over an order field whose positive elements are squares, then $V$ is , or , iff $\operatorname{Gram}[e_{1},\ldots,e_{n}]=0$.

## References

Title Gram determinant GramDeterminant 2013-03-22 15:41:37 2013-03-22 15:41:37 CWoo (3771) CWoo (3771) 13 CWoo (3771) Definition msc 15A63 GrammianDeterminant GramMatrix