We fix a real vector space $E$ of dimension $n$, as well as a nilpotent vector subspace $\mathcal{V}$ of $\mathcal{L}(E)$, equipped with an inner product $(-\mid-)$. We consider an arbitrary vector $x$ of $E \backslash \{0\}$, and set
$$H := \operatorname{Vect}(x)^{\perp}, \quad \mathcal{V} x := \{v(x) \mid v \in \mathcal{V}\} \text{ and } \mathcal{W} := \{v \in \mathcal{V} : v(x) = 0\}$$
We denote by $\pi$ the orthogonal projection of $E$ onto $H$. For $u \in \mathcal{W}$, we denote by $\bar{u}$ the endomorphism of $H$ defined by $\forall z \in H, \bar{u}(z) = \pi(u(z))$. We consider the sets $\overline{\mathcal{V}} := \{\bar{u} \mid u \in \mathcal{W}\}$ and $\mathcal{Z} := \{u \in \mathcal{W} : \bar{u} = 0\}$. Given $a \in E$ and $x \in E$, $(a \otimes x)(z) = (a \mid z) \cdot x$ for all $z \in E$.
Show that there exists a vector subspace $L$ of $E$ such that
$$\mathcal{Z} = \{a \otimes x \mid a \in L\} \quad \text{and} \quad \operatorname{dim} L = \operatorname{dim} \mathcal{Z},$$
and show that then $x \in L^{\perp}$.