I'm struggling in proving collinearity of two vectors provided that the relations described here below hold. Also, I was wondering if such condition can be relaxed.
Consider a vector $x\in\mathbb{R}^n$ and the matrices $J_i\in\mathbb{R}^{n\times n}$, $i=1\dots n-1$. $J_i$ are skew-symmetric matrices, i.e. $J_i+J_i^\top=0$. Define then the matrix $$ M:=\begin{bmatrix} x^\top J_i\\ \vdots\\ x^\top J_{n-1}\\ \end{bmatrix}\in\mathbb{R}^{(n-1)\times n}, $$ and let $y\in\mathbb{R}^n$ such that $My=0$.
Prove that if (and only if?) $\operatorname{rank}(M)=n-1$ then $x$ and $y$ are collinear, i.e. there exists a scalar $\gamma$ such that $x=\gamma y$.
Of course, the result is trivial if $x = 0,$ hence we may assume that $x$ and $y$ are both nonzero. Observe that $M$ determines a linear transformation $T : \mathbb R^n \to \mathbb R^{n - 1},$ hence by assumption that $\operatorname{rank}(T) = \operatorname{rank}(M) = n - 1,$ it follows by the Rank-Nullity Theorem that $\ker(T)$ has dimension $1.$ Considering that $My = 0,$ it follows that $\ker(T)$ is generated by the nonzero vector $y,$ and it suffices to show that $Mx = 0.$ Observe that $$Mx = \begin{bmatrix} x^t J_1 x \\ x^t J_2 x \\ \vdots \\ x^t J_{n - 1} x \end{bmatrix}.$$ By the result proven here (or here, if you prefer), it follows that $Mx = 0.$ QED.