[proofplan] We first verify that the assignment does not depend on the representative of the projective class $[\ell]$ and that $\ker \ell$ is a codimension-one subspace of $V$. Then we prove injectivity by showing that two nonzero linear functionals with the same projectivized kernel have the same kernel, hence differ by a nonzero scalar. Finally, we prove surjectivity by constructing, from any codimension-one subspace $W \leq V$, a nonzero linear functional whose kernel is exactly $W$. [/proofplan]

[step:Define the assignment and verify that its value is a projective hyperplane]Define \begin{align*} \Phi: \mathbb{P}(V^*) \to \{\mathbb{P}(W) : W \leq V \text{ and } \dim_K W = \dim_K V - 1\} \end{align*} by \begin{align*} \Phi([\ell]) = \mathbb{P}(\ker \ell), \end{align*} where $\ell \in V^*$ is any nonzero representative of the projective class $[\ell]$. If $\ell' = \lambda \ell$ for some $\lambda \in K^\times$, then \begin{align*} \ker \ell' = \{v \in V : \lambda \ell(v) = 0\} = \{v \in V : \ell(v) = 0\} = \ker \ell. \end{align*} Thus $\Phi([\ell])$ is independent of the chosen nonzero representative. It remains in this step to check that $\mathbb{P}(\ker \ell)$ is a projective hyperplane. Since $\ell \neq 0$, choose $u \in V$ with $\ell(u) \neq 0$. For every $v \in V$, define \begin{align*} a_v = \frac{\ell(v)}{\ell(u)} \in K. \end{align*} Then $v - a_v u \in \ker \ell$, because \begin{align*} \ell(v - a_v u) = \ell(v) - a_v \ell(u) = 0. \end{align*} Hence every $v \in V$ has a decomposition \begin{align*} v = (v - a_v u) + a_v u \end{align*} with $v - a_v u \in \ker \ell$ and $a_v u \in Ku$. Also $\ker \ell \cap Ku = \{0\}$, since if $bu \in \ker \ell$, then \begin{align*} 0 = \ell(bu) = b\ell(u), \end{align*} and $\ell(u) \neq 0$ implies $b = 0$. Therefore \begin{align*} V = \ker \ell \oplus Ku. \end{align*} Since $Ku$ is one-dimensional, $\dim_K \ker \ell = \dim_K V - 1$. Thus $\mathbb{P}(\ker \ell)$ is a projective hyperplane in $\mathbb{P}(V)$.[/step]

[guided]We must check two things before discussing bijectivity. First, the formula must not depend on the representative of the point $[\ell] \in \mathbb{P}(V^*)$. By the quotient description of [projective space](/page/Projective%20Space) in [citetheorem:9466], two nonzero functionals $\ell,\ell' \in V^*$ represent the same point precisely when $\ell' = \lambda \ell$ for some $\lambda \in K^\times$. For such a scalar $\lambda$, we have \begin{align*} \ker \ell' = \{v \in V : \ell'(v) = 0\} = \{v \in V : \lambda \ell(v) = 0\}. \end{align*} Because $\lambda \neq 0$, the equation $\lambda \ell(v) = 0$ is equivalent to $\ell(v) = 0$. Hence \begin{align*} \ker \ell' = \ker \ell. \end{align*} So $\Phi([\ell]) = \mathbb{P}(\ker \ell)$ is independent of the representative. Second, we must show that $\ker \ell$ has codimension one. Since $\ell$ is nonzero as an element of $V^*$, there exists $u \in V$ such that $\ell(u) \neq 0$. For each $v \in V$, define the scalar \begin{align*} a_v = \frac{\ell(v)}{\ell(u)}. \end{align*} Then \begin{align*} \ell(v - a_v u) = \ell(v) - a_v \ell(u) = \ell(v) - \ell(v) = 0. \end{align*} Thus $v - a_v u \in \ker \ell$, and consequently \begin{align*} v = (v - a_v u) + a_v u \end{align*} is a sum of an element of $\ker \ell$ and an element of the one-dimensional subspace $Ku$. The sum is direct: if $bu \in \ker \ell$ for some $b \in K$, then \begin{align*} 0 = \ell(bu) = b\ell(u). \end{align*} Since $\ell(u) \neq 0$, this forces $b = 0$, so $\ker \ell \cap Ku = \{0\}$. Therefore \begin{align*} V = \ker \ell \oplus Ku. \end{align*} The subspace $Ku$ has dimension $1$, so $\ker \ell$ has dimension $\dim_K V - 1$. Therefore $\mathbb{P}(\ker \ell)$ is exactly a projective hyperplane under the stated convention that projective hyperplanes are projectivizations of codimension-one linear subspaces.[/guided]

[step:Show that equal projectivized kernels force proportional functionals] Let $[\ell],[m] \in \mathbb{P}(V^*)$ and suppose \begin{align*} \Phi([\ell]) = \Phi([m]). \end{align*} Then \begin{align*} \mathbb{P}(\ker \ell) = \mathbb{P}(\ker m). \end{align*} Since $\dim_K V \geq 2$ and each kernel has dimension $\dim_K V - 1$, both $\ker \ell$ and $\ker m$ are nonzero subspaces. We claim that $\ker \ell = \ker m$. If $v \in \ker \ell$ and $v \neq 0$, then $[v] \in \mathbb{P}(\ker \ell) = \mathbb{P}(\ker m)$, so there are $w \in \ker m \setminus \{0\}$ and $\lambda \in K^\times$ such that $v = \lambda w$. Hence $v \in \ker m$. Since $0 \in \ker m$, this proves $\ker \ell \subset \ker m$. Interchanging $\ell$ and $m$ gives $\ker m \subset \ker \ell$, so \begin{align*} \ker \ell = \ker m. \end{align*} Choose $u \in V$ with $\ell(u) \neq 0$. Since $\ker \ell = \ker m$, this $u$ is not in $\ker m$, so $m(u) \neq 0$. Define \begin{align*} c = \frac{m(u)}{\ell(u)} \in K^\times. \end{align*} For any $v \in V$, define \begin{align*} a_v = \frac{\ell(v)}{\ell(u)} \in K. \end{align*} Then $v - a_v u \in \ker \ell = \ker m$, and hence \begin{align*} m(v) = m(a_v u) = a_v m(u) = \frac{\ell(v)}{\ell(u)}m(u) = c\ell(v). \end{align*} Therefore $m = c\ell$, with $c \in K^\times$, so $[m] = [\ell]$ in $\mathbb{P}(V^*)$. Thus $\Phi$ is injective. [/step]

[step:Construct a functional with a prescribed codimension-one kernel]Let $\mathbb{P}(W)$ be a projective hyperplane in $\mathbb{P}(V)$. By definition of the target set, $W \leq V$ and \begin{align*} \dim_K W = \dim_K V - 1. \end{align*} Choose a basis $(w_1,\ldots,w_{n-1})$ of $W$, where $n = \dim_K V$. Since $\dim_K W = n - 1$, extend this basis to a basis $(w_1,\ldots,w_{n-1},u)$ of $V$. Define a [linear map](/page/Linear%20Map) \begin{align*} \ell: V \to K \end{align*} by prescribing its values on this basis: \begin{align*} \ell(w_i) = 0 \text{ for } 1 \leq i \leq n - 1, \qquad \ell(u) = 1. \end{align*} The basis prescription determines a unique $K$-linear functional $\ell \in V^*$. Since $\ell(u) = 1$, we have $\ell \neq 0$. Also $W \subset \ker \ell$ by construction. Conversely, if $v \in \ker \ell$, write uniquely \begin{align*} v = a_1w_1 + \cdots + a_{n-1}w_{n-1} + bu \end{align*} with $a_1,\ldots,a_{n-1},b \in K$. Applying $\ell$ gives \begin{align*} 0 = \ell(v) = b. \end{align*} Hence $v \in W$. Therefore $\ker \ell = W$, and so \begin{align*} \Phi([\ell]) = \mathbb{P}(\ker \ell) = \mathbb{P}(W). \end{align*} Thus $\Phi$ is surjective.[/step]

[guided]We now start with an arbitrary projective hyperplane and build the point of the dual projective space that maps to it. Let $\mathbb{P}(W)$ be a projective hyperplane. This means that $W$ is a linear subspace of $V$ of codimension one, so if $n = \dim_K V$, then \begin{align*} \dim_K W = n - 1. \end{align*} Choose a basis $(w_1,\ldots,w_{n-1})$ of $W$. Since $V$ has dimension $n$, this basis can be extended to a basis $(w_1,\ldots,w_{n-1},u)$ of $V$ for some vector $u \in V$. The idea is to define a linear functional that vanishes on the hyperplane direction $W$ and is nonzero in the one remaining direction. Define \begin{align*} \ell: V \to K \end{align*} by requiring \begin{align*} \ell(w_i) = 0 \text{ for } 1 \leq i \leq n - 1, \qquad \ell(u) = 1. \end{align*} A function defined on a basis extends uniquely to a linear map on the whole [vector space](/page/Vector%20Space), so this prescription defines a unique element $\ell \in V^*$. Because $\ell(u) = 1$, the functional is nonzero, and hence $[\ell]$ is a valid point of $\mathbb{P}(V^*)$. We verify that its kernel is exactly $W$. First, each basis vector $w_i$ of $W$ satisfies $\ell(w_i) = 0$, so every linear combination of the $w_i$ also lies in $\ker \ell$. Thus \begin{align*} W \subset \ker \ell. \end{align*} For the reverse inclusion, let $v \in \ker \ell$. Since $(w_1,\ldots,w_{n-1},u)$ is a basis of $V$, there are unique scalars $a_1,\ldots,a_{n-1},b \in K$ such that \begin{align*} v = a_1w_1 + \cdots + a_{n-1}w_{n-1} + bu. \end{align*} Applying $\ell$ and using linearity gives \begin{align*} 0 = \ell(v) = a_1\ell(w_1) + \cdots + a_{n-1}\ell(w_{n-1}) + b\ell(u) = b. \end{align*} Therefore \begin{align*} v = a_1w_1 + \cdots + a_{n-1}w_{n-1} \in W. \end{align*} So $\ker \ell \subset W$, and hence $\ker \ell = W$. Consequently \begin{align*} \Phi([\ell]) = \mathbb{P}(\ker \ell) = \mathbb{P}(W). \end{align*} Because the hyperplane $\mathbb{P}(W)$ was arbitrary, every projective hyperplane lies in the image of $\Phi$.[/guided]

[step:Conclude that the correspondence is a bijection] The previous steps show that $\Phi$ is well-defined, injective, and surjective. Therefore \begin{align*} \Phi: \mathbb{P}(V^*) \to \{\mathbb{P}(W) : W \leq V \text{ and } \dim_K W = \dim_K V - 1\} \end{align*} is a bijection. Equivalently, projective hyperplanes in $\mathbb{P}(V)$ are naturally parametrized by points of the dual projective space $\mathbb{P}(V^*)$. [/step]