Gram matrix invertible iff set of vectors linearly independent Given a set of vectors $v_1 \cdots v_n$, the $n\times n$ Gram matrix $G$ is defined as
$G_{i,j}=v_i \cdot v_j$
Due to symmetry in the dot product, $G$ is Hermitian.
I'm trying to remember why $|G|=0$ iff the set of vectors are not linearly independent.

JasonMond's "only if" part is not as general as it should, because s/he assumes that $A$ is square.  In the following, I complete the proof that holds whether $A$ is square or not.
Let $G = A^T A$.  If the column vectors of $A$ are linearly dependent, there exists
a vector $u \neq 0$ such that
$$
  A u = 0.
$$
It follows that
$$
  0 = A^T A u = G u.
$$
Since $u \neq 0$, $G$ is not invertible.
Conversely, if $G$ is not invertible, there exists a vector $v \neq 0$ such that
$$
  G v = 0.
$$
It follows that
$$
  0 = v^T G v = v^T A^T A v = (A v)^T A v
    = \lVert A v\rVert^2
$$
and therefore that
$$
   A v = 0.
$$
Since $v \neq 0$, the column vectors of $A$ are linearly dependent.  QED.
