Let $m \geq 2$ be a natural integer and $E$ an $\mathbb{R}$-vector space of dimension $2m+1$ equipped with a scalar product $(.|.)$. Let $T$ be an endomorphism of $E$ satisfying (H1): $T^{2m} \neq 0_{\mathcal{L}(E)}$ and $T^{2m+1} = 0_{\mathcal{L}(E)}$. We consider the map $S$ from $E \times E$ to $\mathbb{R}$ defined by $$\forall (v,w) \in E^2, S(v,w) = (v \mid T(w)) + (T(v) \mid w)$$ and we denote by $G$ the set of elements $u \in E$ satisfying: (a) $u \in \operatorname{Im}(T)$, (b) $\forall v \in E, S(u,v) = 0$. Show that $G$ is a vector subspace of $E$ and that $G \cap \operatorname{ker}(T) = \{0_E\}$.
Let $m \geq 2$ be a natural integer and $E$ an $\mathbb{R}$-vector space of dimension $2m+1$ equipped with a scalar product $(.|.)$. Let $T$ be an endomorphism of $E$ satisfying (H1): $T^{2m} \neq 0_{\mathcal{L}(E)}$ and $T^{2m+1} = 0_{\mathcal{L}(E)}$. We consider the map $S$ from $E \times E$ to $\mathbb{R}$ defined by
$$\forall (v,w) \in E^2, S(v,w) = (v \mid T(w)) + (T(v) \mid w)$$
and we denote by $G$ the set of elements $u \in E$ satisfying:
(a) $u \in \operatorname{Im}(T)$,
(b) $\forall v \in E, S(u,v) = 0$.
Show that $G$ is a vector subspace of $E$ and that $G \cap \operatorname{ker}(T) = \{0_E\}$.