Why for any $(n-l+1)$ orthogonal unit vectors $\{v_1,...v_{n-l+1}\}$ ,$ \sum\limits_{i=1}^{n-l+1}a_{\alpha\beta}v^\alpha_iv^\beta_i\ge0 $

Question

Let $V$ be a n-dimensional vector space,$A=\{a_{\alpha\beta}\} $ is a $n\times n$ antisymmetric matrix ($A^T=-A$), and $rank A=l$.

Why for any $(n-l+1)$ orthogonal unit vectors $\{v_1,...v_{n-l+1}\}$ $$ \sum\limits_{i=1}^{n-l+1}a_{\alpha\beta}v^\alpha_iv^\beta_i\ge0 ~~~~~~~~~~\alpha~\beta ~~\text{are summed up} $$

Answer 1

It is false.

Let $ \ V = \mathbb{R}^3$, $A = \left( \begin{array}{ccc} -1 & 0 & 1 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{array} \right)$, $v_1 = (1,0,0) \ $ and $ \ v_2 = (0,1,0)$. Then $ \ n=3$, $Av_1 = -v_1,$ $Av_2 = -v_2$, $\ell = rank(A)=2$, $n- \ell +1 = 2 \ $ and $$\sum\limits_{i=1}^{n-l+1}a_{\alpha\beta}v^\alpha_iv^\beta_i = Av_1 \bullet v_1 + Av_2 \bullet v_2 = (-1) + (-1) = -2 <0$$