[proofplan]
Choose a nonzero linear form defining each hyperplane in the central arrangement. These forms define a representable matroid, and the key point is that intersections of hyperplanes are exactly annihilators of spans of the corresponding defining forms. Passing from a set of forms to its matroid closure does not change the intersection, so the elements of the intersection lattice correspond precisely to flats of this matroid. The order convention on $L(\mathcal A)$ is reverse inclusion, which makes this correspondence an order isomorphism with the lattice of flats; the result then follows from the standard theorem that flat lattices of finite matroids are geometric.
[/proofplan]
[step:Choose defining linear forms and form the associated representable matroid]
For each hyperplane $H\in\mathcal A$, choose a nonzero linear functional $\alpha_H\in V^*$ such that
\begin{align*}
H=\ker(\alpha_H).
\end{align*}
This is possible because each $H$ is a linear hyperplane in the finite-dimensional [vector space](/page/Vector%20Space) $V$.
Let $E:=\mathcal A$ be the finite indexing set of hyperplanes. Define a rank function
\begin{align*}
r:2^E&\to \mathbb N\cup\{0\}
\end{align*}
\begin{align*}
S&\mapsto \dim_k\operatorname{span}_k\{\alpha_H:H\in S\}.
\end{align*}
The finite set of vectors $(\alpha_H)_{H\in E}$ in the vector space $V^*$ defines the representable matroid $M$ on ground set $E$ with this rank function. For a subset $S\subset E$, its closure in $M$ is
\begin{align*}
\operatorname{cl}_M(S)=\{H\in E:\alpha_H\in \operatorname{span}_k\{\alpha_K:K\in S\}\}.
\end{align*}
A flat of $M$ is a subset $F\subset E$ satisfying $\operatorname{cl}_M(F)=F$.
[/step]
[step:Express each hyperplane intersection as an annihilator]
For every subset $S\subset E$, define the linear subspace
\begin{align*}
W_S:=\operatorname{span}_k\{\alpha_H:H\in S\}\subset V^*
\end{align*}
and define its annihilator in $V$ by
\begin{align*}
W_S^\circ:=\{v\in V:\beta(v)=0\text{ for every }\beta\in W_S\}.
\end{align*}
Then
\begin{align*}
W_S^\circ=\bigcap_{H\in S}H.
\end{align*}
Indeed, if $v\in W_S^\circ$, then $\alpha_H(v)=0$ for every $H\in S$, so $v\in H$ for every $H\in S$. Conversely, if $v\in\bigcap_{H\in S}H$, then $\alpha_H(v)=0$ for every $H\in S$, and hence every $k$-linear combination of the $\alpha_H$ also vanishes on $v$; therefore $v\in W_S^\circ$.
[/step]
[step:Show that matroid closure does not change the intersection]
For every subset $S\subset E$, we have
\begin{align*}
\bigcap_{H\in S}H=\bigcap_{H\in \operatorname{cl}_M(S)}H.
\end{align*}
Since $S\subset \operatorname{cl}_M(S)$, the right-hand intersection is contained in the left-hand intersection. For the reverse inclusion, let $v\in\bigcap_{H\in S}H$. If $K\in\operatorname{cl}_M(S)$, then $\alpha_K\in W_S$, so $\alpha_K(v)=0$ by the annihilator description above. Thus $v\in K$ for every $K\in\operatorname{cl}_M(S)$, giving the reverse containment.
[guided]
The purpose of this step is to replace arbitrary subsets of hyperplanes by flats of the matroid without changing the element of the intersection lattice. Fix a subset $S\subset E$. The inclusion
\begin{align*}
S\subset \operatorname{cl}_M(S)
\end{align*}
means that intersecting over $\operatorname{cl}_M(S)$ imposes at least as many hyperplane equations as intersecting over $S$. Therefore
\begin{align*}
\bigcap_{H\in \operatorname{cl}_M(S)}H\subset \bigcap_{H\in S}H.
\end{align*}
For the opposite inclusion, take $v\in\bigcap_{H\in S}H$. This means $\alpha_H(v)=0$ for every $H\in S$. Since $K\in\operatorname{cl}_M(S)$ means
\begin{align*}
\alpha_K\in\operatorname{span}_k\{\alpha_H:H\in S\},
\end{align*}
there exist finitely many scalars $c_H\in k$, indexed by $H\in S$, such that
\begin{align*}
\alpha_K=\sum_{H\in S}c_H\alpha_H.
\end{align*}
Evaluating at $v$ gives
\begin{align*}
\alpha_K(v)=\sum_{H\in S}c_H\alpha_H(v)=0.
\end{align*}
Thus $v\in K$. Since this holds for every $K\in\operatorname{cl}_M(S)$, we obtain
\begin{align*}
v\in\bigcap_{K\in\operatorname{cl}_M(S)}K.
\end{align*}
Hence
\begin{align*}
\bigcap_{H\in S}H=\bigcap_{H\in \operatorname{cl}_M(S)}H.
\end{align*}
This is the precise point where matroid closure enters: adding all hyperplanes whose defining forms are forced by the linear span of the original defining forms adds no new geometric condition on the intersection.
[/guided]
[/step]
[step:Identify the intersection lattice with the lattice of flats]
Let $\mathcal F(M)$ denote the set of flats of the matroid $M$, ordered by inclusion. Define
\begin{align*}
\Phi:\mathcal F(M)&\to L(\mathcal A)
\end{align*}
\begin{align*}
F&\mapsto \bigcap_{H\in F}H.
\end{align*}
The map $\Phi$ is surjective because every element of $L(\mathcal A)$ has the form $\bigcap_{H\in S}H$, and the previous step gives
\begin{align*}
\bigcap_{H\in S}H=\bigcap_{H\in \operatorname{cl}_M(S)}H
\end{align*}
with $\operatorname{cl}_M(S)$ a flat.
To prove injectivity, let $F,G\in\mathcal F(M)$ and suppose $\Phi(F)=\Phi(G)$. Then
\begin{align*}
W_F^\circ=W_G^\circ.
\end{align*}
For a subspace $W\subset V^*$, finite-dimensional linear algebra gives
\begin{align*}
(W^\circ)^\perp=W,
\end{align*}
where
\begin{align*}
Y^\perp:=\{\beta\in V^*:\beta(y)=0\text{ for every }y\in Y\}
\end{align*}
for a subspace $Y\subset V$. Applying this identity to $W_F$ and $W_G$ yields $W_F=W_G$. Therefore
\begin{align*}
F=\{H\in E:\alpha_H\in W_F\}=\{H\in E:\alpha_H\in W_G\}=G,
\end{align*}
because $F$ and $G$ are flats. Hence $\Phi$ is bijective.
Finally, if $F\subset G$ are flats, then
\begin{align*}
\bigcap_{H\in G}H\subset \bigcap_{H\in F}H.
\end{align*}
Since $L(\mathcal A)$ is ordered by reverse inclusion, this says precisely that
\begin{align*}
\Phi(F)\le \Phi(G)
\end{align*}
in $L(\mathcal A)$. Thus $\Phi$ is an order isomorphism from the flat lattice of $M$ to $L(\mathcal A)$.
[/step]
[step:Conclude geometricity from the flat lattice theorem]
The matroid $M$ is finite because its ground set $E=\mathcal A$ is finite. By [citetheorem:8112], the lattice of flats of a finite matroid is a geometric lattice. Since $L(\mathcal A)$ is order-isomorphic to the flat lattice of $M$, the lattice $L(\mathcal A)$ is also geometric. This proves the theorem.
[/step]
