# Exercises on hyperplane and applications

In this article, we propose exercises on hyperplane and applications. For instance, a hyperplane is a vector subspace of dimension equal to the dimension of the whole space minus one.

### Exercises on hyperplane

Exercise: Let $V$ be a vector space and $n\in\mathbb{N}$, let $H$ be a hyperplane of $V$, and let $v\in V$ be a vector. Under what condition the subspaces $H$ and ${\rm span}(v):=\{\lambda v:\lambda\in\mathbb{C}\}$ are supplementally in $V$.

Solution: We shall discuss two cases:

First case: if $a\in H$. Then for any $\lambda\in\mathbb{C},$ $\lambda a\in H$. Thus ${\rm span}(v)\subset H,$ so that $H$ and ${\rm span}(v)$ are not supplementally.

Second case: if $a\notin H$. First of all, we have $\dim({\rm span}(v))=1$. As $H$ is a hyperplane of $V,$ it follows that $\dim(H)=n-1$. Hence \begin{align*} \dim(H)+\dim({\rm span}(v))=n=\dim(V). \end{align*} Now let $x\in H\cap {\rm span}(v)$. Then $x\in H$ and there exists $\lambda\in\mathbb{C}$ such that $x=\lambda a$. We necessarily have $\lambda=0,$ because if not, then $a=\lambda^{-1} x\in H,$ absurd. Hence $x=0_V$, and then $H\cap {\rm span}(v)=\{0_V\}$. This show that $H+{\rm span}(v)=V$ is a direct sum.

Exercise: Let $\psi: \mathbb{R}^n\to \mathbb{C}$ be a non null linear forme and $\Phi$ be an endomorphism of $\mathbb{R}^n$. Prove that the kernel of $\psi$ is stable by $\Phi$ (i.e. $\Phi(\ker(\psi))\subset \ker(\psi)$) if and only if there exists a real $\lambda\in\mathbb{R}$ such that $\psi\circ\Phi=\lambda\psi$.

Solution: Assume that there exists $\lambda\in\mathbb{R}$ such that $\psi\circ\Phi=\lambda\psi$. Let $x\in \ker(\psi)$ and prove that $\Phi(x)\in \ker(\psi)$. In fact, we have $\psi(x)=0$. On the other hand, \begin{align*} \psi(\Phi(x))=\lambda \psi(x)=0. \end{align*} This implies that $\Phi(\ker(\psi))\subset \ker(\psi)$.

Conversely, assume that $\ker(\psi)$ is stable by $\Phi$. Observe that if $x\in \ker(\psi)$ then $\psi(\Phi(x))=0,$ so that \begin{align*} \psi(\Phi(x))=\lambda \psi(x),\quad \forall x\in \ker(\psi),\;\forall \lambda\in\mathbb{R}. \end{align*} It suffice then to look for real $\lambda$ and a supplementally space $K$ of $\ker(\psi)$ such that $\psi\circ\Phi=\lambda\psi$ on $K$. According to rank theorem we have $\ker(\psi)$ is a hyperplane, so $\dim(\ker(\psi))=n-1$. Thus any supplementally space $K$ of $\ker(\psi)$ will satisfies $\dim(K)=1$. Take then $a\in \mathbb{R}^n$ such that $\psi(a)\neq 0$ (a such $a$ exists because $\psi$ is a non null forme), so that $a\notin \ker{\psi}$. This implies that ${\rm span}(a)\cap \ker(\psi)=\{0\}$. As $\dim({\rm span}(a))=1$ and $\dim(\ker{\psi})+\dim({\rm span}(a))=n$. Then \begin{align*} \ker(\psi)\oplus {\rm span}(a)=\mathbb{R}^n. \end{align*} Then it suffices to look for real $\lambda$ such that $\psi\circ\Phi=\lambda\psi$ on ${\rm span}(a)$. In particular $\psi(\Phi(a))=\lambda \psi(a)$. We choose \begin{align*} \lambda=\frac{\psi(\Phi(a))}{\psi(a)}. \end{align*}