# Determinant of Gram matrix - without knowing it's a Gram matrix or using anything “fancy” [closed]

NOTE this is independent learning or self learning.

Hey I hate to "dumb the question down" but I am trying to find an alternate proof that doesn't use knowledge of what a Gram matrix is, imagine this just happens to be one.

Let $G(v_1,v_2,...,v_k)$ the Gram matrix which (as I can't do matrices in latex) is the matrix where $A_{i,j}=<v_i,v_j>$ where $<,>$ is an inner product.

This makes the matrix symmetric.

I wish to prove:

If ${v_1,v_2,...,v_k}$ are linearly dependent then $\det(G(v_1,v_2,..,v_k))=0$

and

if ${v_1,v_2,...,v_k}$ are linearly independent then $\det(G(v_1,...,v_k))\ne 0$ (or $>0$ as the matrix has entirely positive members).

I believe that this can be done from definitions because if one considers the case k=2, if the points lie on a line going through the origin then this determinant is zero.

I know the "Schwarz" inequality, $\lvert\sum^N_{i=1}x_iy_i\rvert\le(\sum^N_{i=1}(x_i)^2)^\frac{1}{2}(\sum^N_{i=1}(y_i)^2)^\frac{1}{2}$, I also proved if this is an equality (not less than) then either all the $x_i$ are zero, or $y_i=\lambda x_i$ for all i

$\lambda\boldsymbol{v}$ = line through the orgin as one varies lambda.

Please no help that uses definitions of the Gram matrix, if this doesn't rule out that $A^TA$ method, please rule it out because I'd like to use the definition of $<,>$ rather than using a dot product.

• It strikes me as very arrogant to say "I have no idea how this helps" and request an answerer move their hint to a comment within mere minutes of seeing their response, surely not enough time to try to take full advantage of the hint and see if it truly helps or not, even if it is just a sentence or two. Note $B^TB=A$ in Jagy's answer is true regardless of whether or not $B$ is square or how many $v$'s there are relative to the dimension of the vector space. – anon Nov 23 '13 at 23:09