Let $A\in\mathbb{R}^{d\times d}$ be a symmetric matrix, and $X_1,\dots, X_n\in \mathbb{R}^d$ be vectors with $n>d$ (if more convenient, one can assume ${\rm span}(X_1,\dots,X_n)=\mathbb{R}^d$.

Assume that, for every $i $, $X_i^T AX_i=0$. What does this tell us about $A$ (of course, as a function of $X_1,\dots,X_n$?

For instance, if we had that $x^T Ax = 0$ for every $x\in\mathbb{R}^d$, then we would have obtained that $A$ is skew-symmetric, that is, $A^T=-A$, which, together with the fact that $A$ is symmetric, would have yielded $A=0$.

Yet another example, take $n=d$, and $X_i=e_i$, the $i ^{th}$ element of the standard basis of $d-$dimensional Euclidean space. Then, all we can say is that the diagonal entries of $A$ are $0$.

If it will make the things simple: Assume that, $n>d^2$, and that, $X_1,\dots,X_n$ are iid random vectors with $d-$iid standard normal entries. What then happens? Can we deduce $A=0$, if conditional on some event, we have more equations than the unknowns?


Here's a partial answer. If $\ n\ge \frac{d\left(d+1\right)} {2}\ $, and the matrices $\ X_i^\top X_i,\ $ $i= 1,2,\dots,n\ $ span the space of symmetric $\ d\times d\ $matrices (which will be true with probability $1$ whenever the entries of the $\ X_i\ $ are independent normal variates with positive variance) then you can conclude that $\ A=0\ $.

This follows from the fact that if the stated condition is true, then for every pair of integers $\ j,k\ $ with $\ 1\le j\le k \le d\ $, there will exist $\ \alpha_1, \alpha_2, \dots, \alpha_n $ such that $\ \sum_\limits{i=1}^n \alpha_i X_i^\top X_i=e_j^\top e_k + e_k^\top e_j \ $. We then have $\ 0= \sum_\limits{i=1}^n \alpha_i X_i^\top AX_i\ = e_j^\top Ae_k + e_k^\top Ae_j= 2a_{jk}\ $, where $\ a_{jk}\ $ is the entry in the $\ j^\mathrm{\,th}\ $ row and $\ k^\mathrm{\,th}\ $ column of $\ A\ $.


@lonza leggiera: many thanks. just for archival purposes, I've decided to fill in several steps of the outline you've provided.

Recall that, $X_1,\dots,X_n \sim N(0,I_d)$ iid, with $X_i \in \mathbb{R}^d$ (namely, $X_i$ to be regarded as a column vector).

1) If $n>d^2$ then with probability $1$, ${\rm span}(X_iX_i^T :1\leqslant i \leqslant n) = S^{d \times d}$, where $S^{d\times d}=\{M\in \mathbb{R}^{d\times d}:M^T=M\}$. To see this, perhaps one way is to argue as follows. Consider an operator $V$ acting on $X_iX_i^T$, and vectorizing it, and consider a $d^2\times n$ matrix $M$ whose $i^{th}$ columns is $X_iX_i^T$. We claim that, $\mathbb{P}[{\rm rank}(M)=d^2]=1$. Indeed, assuming not, we have the event $\{{\rm rank}(M)<d^2\}$. On this event, any $d^2$ vectors form a $d^2\times d^2$ matrix with zero determinant. Let $X$ be a matrix formed by columns $V(X_iX_i^T)$ for $1\leqslant i\leqslant d^2$. Then, $\{{\rm rank}(M)<d^2\}\subseteq \{{\rm det}(X)=0\}$, where the final event clearly has measure $0$, as it is a function of iid continuous random variables.

Suppose therefore that in the remainder, we condition on $\mathcal{E}=\{{\rm span}(X_iX_i^T:1\leqslant i \leqslant n)=S^{d\times d}\}$, where $\mathbb{P}[\mathcal{E}]=1$.

2) Let $e_k,e_\ell$ be $k^{th}$ and $\ell^{th}$ basis vectors for $d-$dimensional Euclidean space. Then, conditional on $\mathcal{E}$, we have that there exists $\theta_i^{(k,\ell)}$ such that: $$ \sum_{i =1}^n \theta_i^{(k,\ell)}X_iX_i^T = e_k e_\ell^T + e_\ell e_k^T. $$

3) Now, observe that, ${\rm trace}(X_i^T A X_i)= {\rm trace}(X_iX_i^T A)$. Hence, $$ 0=\sum_{i =1}^n \theta_i^{(k,\ell)}X_i^T A X_i = \sum_{i =1}^n {\rm trace}(\theta_i^{(k,\ell)} X_i X_i^T A) = {\rm trace}(\sum_{i =1}^n \theta_i^{(k,\ell)} X_i X_i^T A) = {\rm trace}(e_ke_\ell^T A)+{\rm trace}(e_\ell e_k^T A)=2A_{k,\ell}. $$ Therefore, $A_{k,\ell}=0$, for every $1\leqslant k\leqslant \ell \leqslant d$, as claimed.


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