# Link between Gram Matrix and volume of parallelpiped question - Determinant

I'm not sure how to approach this question. I have proved $det(G(v_1,...v_k)≥0$ and have proven the triangle inequality using the Cauchy-Schwarz inequality.

Well, I think the most clean way to proceed here is by induction. Fon $n=1$ we have: $$det(G(v_1))=Vol_1(P(v_1))^2=<v_1,v_1>$$ Now let us suppose the formula valid for $n-1$ and induce the formula for $n$. First of all we notice that both $det(G(v_1,...,v_N))$ and $Vol_n(P(v_n))$ do not depend from a specific choice of the base (that's because the first one is a determinant and the second one is originated by the norm). Now instead of calculating directly $det(G(v_1,...,v_n))$ we calculate the determinant after a change of basis $$det(G(v_1,...,v_n))=det(S^{-1}G(v_1',...,v_n')S)$$ where $v_n'$ is orthogonal to all $v_i'$ with $i<n$. In this coordinates the calculation should be a little more convenient: $$det(G(v_1,...,v_n))=<v_n',v_n'>det(S^{-1}G(v_1',...,v_{n-1}')S)$$ now using the inductive hypotesis $$<v_n',v_n'>det(G(v_1',...,v_{n-1}'))=<v_n',v_n'>Vol_{n-1}(P(v_1',..,v_{n-1}'))^2$$ And finally using the recursive definition of the Volume and using the fact that the volume is invariant for change of basis we obtain $$Vol_{n}(P(v_1',..,v_{n}'))^2=Vol_{n}(P(v_1,..,v_{n}))^2$$
• S is just a matrix for the change of basis. Note that $det(S^{-1}G(v_1',...,v_n')S)=det(S^{-1})det(G(v_1',...,v_n'))det(S)=det(G(v_1',...,v_n'))$
• The last equality holds because $det(S)det(S^{-1}) = det(SS^{-1}) = det(I) = 1$? Nov 24 '15 at 23:41
• It's a way of viewing that, but simply $det(S^{-1})=det(S)^{-1}$ and since $det(S)$ is a real number as $det(G)$ you can switch them and obtain: $det(S)^{-1}det(S)det(G(v_1...v_n))=det(G(v_1...v_n))$