Ytukyg

let $mathcal H$ be a Hilbert space, and let $g_1,...g_nin mathcal H$. let $G(g_1,...g_n)$ be Gram's determinant:

$left|{begin{array}{ccc}
left(g_{1},g_{1}right) & dots & left(g_{1},g_{n}right)\
vdots & ddots & vdots\
left(g_{n},g_{1}right) & dots & left(g_{n},g_{n}right)
end{array}}right|$

write $(h_1,...h_n)$ to be the result of orthogonalisating $(g_1,...g_n)$, without normalizing the vectors. I want to show that
$G(g_1,...g_n)=G(h_1,...h_n)$

It's not hard to see that suffices to show $G(g_1,...g_n)=G(g_1,h_2,g_3,...g_n)$ when $h_2=g_2+alpha g_1$, such that $g_1perp h_2$. but I didn't manage to finish that up. I tried to calculate it directly, but it gets mass.

Try in 2D first. We have $h_2 = g_2 - (g_1,g_2)g_1$. Let $t = (g_1,g_2)$ and lets say its an inner product over $mathbb R$. Let $mathcal G$ be the matrix whose determinant is $G$.
$$
mathcal G(g_1,h_2)=
begin{bmatrix}
|g_1|^2 & (g_1,h_2)\
(h_2,g_1) & |g_2|^2
end{bmatrix}
=
begin{bmatrix}
|g_1|^2 & t-t|g_1|^2\
t-t|g_1|^2 & |g_2|^2-2t^2 + t^2|g_1|^2
end{bmatrix}
$$
This matrix can be obtained from
$$ mathcal G (g_1,g_2) =
begin{bmatrix}
|g_1|^2 & t\
t & |g_2|^2
end{bmatrix}
$$
by subtracting $t$ times row 1 from row 2 and then from this new matrix subtracting $t$ times column 1 from column 2. Both operations used are valid elementary operations represented by matrices of determinant 1, so the determinant is unchanged.

For higher dimensions you see the emergence of terms
begin{align} |h_2|^2 &= |g_2|^2 - 2t^2 +t^2|g_1|^2 quad &(text{ same as before }) \
(h_2,g_i) &= (g_2,g_i) - t(g_1,g_i) & ineq 2end{align}
The original entries of the second row of $mathcal G(g_i)$ would have been $(g_2,g_i)$. If we subtract $t$ times the first row we get
$$ (g_2,g_i) - t(g_1,g_i)$$
as needed, except the diagonal term.
If we then subtract $t$ times the first column, again the same terms appear, and the diagonal term is now exactly as in the $2times 2$ case.