System of homogeneous quadratic equations

Question

I have $2n$ variables (over $\mathbb{C}$) $x_1,...,x_n,y_1,...,y_n$ along with the $\binom{n}{2}$ equations $x_ix_j + y_iy_j = 0$ for $i \neq j$. Is it true that if $n > 5$, namely when $\binom{n}{2} > 2n$, that the only solution to this system of equations is $x_i = y_j = 0$?

Answer 1

Not true for any $n$. Consider the counterexample of arbitrary $x_j$ and $y_j = i\,x_j$ (where $i$ is the imaginary unit), then $x_jx_k+y_jy_k = x_jx_k + i^2 \,x_jx_k = 0$.

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[ EDIT ] Following up on OP's comment below, these are not the only solutions, for example $x_2=x_3=\ldots=x_n=y_2=y_3=\ldots=y_n=0$ is a solution regardless of $x_1,y_1$, and it is a solution in real numbers if $x_1,y_1 \in \mathbb R$. So is $x_3=\ldots=x_n=y_3=\ldots=y_n=0$ with arbitrary $x_1,x_2$ and $y_1=x_1, y_2=-x_2$, for another example.

To characterize the solutions for arbitrary $n \ge 3$ consider the subset of equations:

$$ \begin{align} x_1x_2+y_1y_2=0 \tag{1} \\ x_1x_3+y_1y_3=0 \tag{2} \\ x_2x_3+y_2y_3 = 0 \tag{3} \end{align} $$

Eliminating $x_3$ between $(2)$ and $(3)$:

$$ \require{cancel} x_2(\cancel{x_1x_3}+y_1y_3)-x_1(\cancel{x_2x_3}+y_2y_3)=0 \quad\iff\quad x_2y_1y_3-x_1y_2y_3=0 \tag{4} $$

Eliminating $x_2$ between $(1)$ and $(4)$:

$$ y_1y_3(\cancel{x_1x_2}+y_1y_2) - x_1(\cancel{x_2y_1y_3}-x_1y_2y_3) = 0 \quad\iff\quad y_2y_3(x_1^2+y_1^2) = 0\tag{5} $$

It follows that, for each triplet of indices like $(1,2,3)$, either $y_2y_3=0$ or $y_1 = \pm i x_1$.