Task: Prove that Gram's determinant $\displaystyle G(x_1,\dots, x_n)=0$ if and only if $\displaystyle x_1, \dots, x_k$ are linearly dependent.

So I know that

$\displaystyle G(x_1,\dots, x_n)=\det \begin{vmatrix} \xi( x_1,x_1) & \xi( x_1,x_2) &\dots & \xi( x_1,x_n)\\

\xi( x_2,x_1) & \xi( x_2,x_2) &\dots & \xi( x_2,x_n)\\

\vdots&\vdots&\ddots&\vdots\\

\xi( x_n,x_1) & \xi( x_n,x_2) &\dots & \xi( x_n,x_n)\end{vmatrix}$

(where $\displaystyle \xi$ denotes an inner product) which means that if $\displaystyle G(x_1,\dots, x_n)=0$, then the determinant has to be equal to 0 as well. Why does it happen only with $\displaystyle x_1,\dots, x_k$ being linearly dependent, though?

Also, sorry for the <br> tags in the matrix but TeX converts my \\ signs that way here. Weird...