Let $A$ be an $n\times n$ symmetric matrix and let $l_1,l_2,...,l_{r+s}$ be $r+s$ linearly independent $n\times 1$ vectors such that for all $n\times 1$ vectors $x$ we have
$$x^TAx=(l_1^Tx)^2+...+(l_r^Tx)-(l_{r+1}^Tx)^2-...-(l_{r+s}^Tx)^2$$
We have to prove that $rank(A)=r+s$.
I cannot think of any approach to this problem.
By plugging $e_i$ in $x$ I am getting
$$(A)_{ii}=(l_1^T)_{i1}^2+...+(l_r^T)^2_{i1}-(l_{r+1}^T)^2_{i1}-...-(l_{r+s}^T)^2_{i1}$$
How can I obtain $rank(A)$ from the above equations?
Edit: As @angryavian answered, then A can be written as $$A=\sum_\limits{1\le i \le r} l_il_i^T - \sum_\limits{r+1 \le i \le s} l_il_i^T$$
we know that if $A,B$ be $n \times n$ symmetric matrix with $x^TAx=x^TBx$ for all $x \in \mathbb{R}^n$ then $A=B$.
Using the above information it remains to show that $$\sum_\limits{1\le i \le r} l_il_i^T - \sum_\limits{r+1 \le i \le s} l_il_i^T$$ has rank $r+s$.
How do I show this?
Hint: Note that $(l_i^\top x)^2 = x^\top l_i l_i^\top x$. Does this suggest a way to write $A$ in terms of the $l_i$?
Hint: Show that $A=LL^\top$ for a certain $n \times (r+s)$ matrix with rank $r+s$.