(Conjectured) Taylor Expansion for Determinant of a Gram Matrix Question: I have a formula that seems plausible and I can "empirically" verify it. Is this something that is already known / does anyone have any ideas on how to prove it? (Motivation is from random matrix theory).
Setting: $k$ unit vectors $v_1,\dots,v_k$ drawn from $\mathbb{R}^n$, with $k \ll n$ and $|\langle v_i, v_j\rangle| < \epsilon $ for all $i \neq j$. Let $V$ be the $k \times n$ matrix with rows $v_i$. We try to understand the determinant, $\det(G)$, of the Gram matrix $G = V\,V^T$.
Guess: for $\epsilon$ sufficiently small, and some error term (maybe $k^2\epsilon^2$?)
$$
\det(G) \stackrel{?}\le \prod_{i=1}^k(1 - \sum_{j = 1}^{k-1} |\langle v_i, v_j\rangle|^2 +  O(\dots))
$$
This seems like some kind of first order Taylor expansion? Numerically, this holds for iid random unit vectors. If it matters, I am interested in the scaling $\epsilon \approx 1/\sqrt{n}$. I know that if $\epsilon \ll 1/n$, this problem trivializes a bit.
 A: This seems to be a direct consequence of the structure of the Gram matrix. Since all the vectors involved are normalized, we can decompose the matrix into a diagonal and off-diagonal part as follows:
$$G_{ij}=v_{i}^T v_j=\delta_{ij}+(1-\delta_{ij})(v_i \cdot v_j)$$
By utilizing the representation of the determinant in terms of the Levi-Civita tensor (Einstein summation convention is implied here),
$$\det(G)=\epsilon_{i_1...i_k}G_{1i_1}...G_{ki_k}$$
we can substitute the decomposition we have above and sort the $2^k$ summands in terms of the number of inner products they involve, which in your notation could be thought of as an "expansion" in powers of $\epsilon$. We see that the determinant can be decomposed in a sum of $k$ terms
$$\det(G)=\sum_{m=0}^kg_m$$
where the first few sums read
$$g_0=\epsilon_{i_1...i_k}\delta_{1i_1}...\delta_{ki_k}=1\\
g_1=\epsilon_{i_1...i_k}[(1-\delta_{1i_1})(v_1\cdot v_{i_1})\delta_{2i_2}...\delta_{ki_k}+...+\delta_{1i_1}...\delta_{(k-1)i_{k-1}}(1-\delta_{ki_k})(v_k\cdot v_{i_k}))\\
g_2=\epsilon_{i_1...i_k}\sum_{m=1}^{k}\sum_{n<m}(1-\delta_{n i_n})(v_n\cdot v_{i_n})(1-\delta_{m i_m})(v_m\cdot v_{i_m})\prod_{\ell\neq m,n}\delta_{\ell i_\ell}$$
Clearly, $g_1=0$ , because in each individual term the $k-1$ Kronecker deltas force the value of $i_m ~,~1\leq m \leq k$ to be equal to $m$, or else the Levi-Civita symbol vanishes, however in that case $(1-\delta_{mi_m})$ vanishes and hence every term is zero. Computing $g_2$ is similarly simple, just write out each term (which is basically a sum over all distinct pairs of inner products) and notice that
$$\epsilon_{i_1...i_n}\delta_{1i_1}...(1-\delta_{q_{i_q}})(v_q\cdot v_{i_q})...(1-\delta_{r_{i_r}})(v_r\cdot v_{i_r})...\delta_{ki_k}=\\ \epsilon_{1...i_q...i_r...k}(1-\delta_{q_{i_q}})(v_q\cdot v_{i_q})(1-\delta_{r_{i_r}})(v_r\cdot v_{i_r})=\\
\epsilon_{1...r...q...k}(v_{q}\cdot v_{r})^2=-(v_{q}\cdot v_{r})^2$$
in step 3, in order for the epsilon symbol to not vanish, we must have $\{i_q, i_r\}\in\{q,r\}$. Of course, the contributions vanish when $i_q=q, i_r=r$ and we are only left with the choice $i_q=r, i_r=q$, which also reverses the sign of the symbol. Hence
$$g_2=-\sum_{q=1}^k\sum_{r<q}(v_{q}\cdot v_r)^2$$
which shows exactly what we wanted to show.
