[proofplan] We first identify the [characteristic polynomial](/page/Characteristic%20Polynomial) of the lattice of flats with the usual subset expansion for the matroid: \begin{align*} \chi_{L(M)}(t)=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}. \end{align*} Then we cancel all subsets containing a broken circuit by a rank-preserving sign-reversing involution. The only surviving subsets are the no-broken-circuit subsets, and these are independent, so their rank equals their cardinality. Comparing coefficients of $t^{r-i}$ gives the stated formula for $w_i$. [/proofplan]

[step:Derive the subset expansion from the flat lattice Möbius function]The closure map declared in the theorem statement is \begin{align*} \operatorname{cl}_M:2^E&\to L(M) \end{align*} \begin{align*} A&\mapsto \operatorname{cl}_M(A). \end{align*} Since $M$ is loopless, $\operatorname{cl}_M(\varnothing)=\varnothing$, so $\varnothing$ is the minimum element of $L(M)$. Define the integer-valued map \begin{align*} S:L(M)&\to \mathbb Z \end{align*} \begin{align*} F&\mapsto \sum_{\substack{A\subset E, \operatorname{cl}_M(A)=F}}(-1)^{|A|}. \end{align*} If $F\in L(M)$, then the flats $G\subset F$ are exactly the possible closures of subsets $A\subset F$. Hence \begin{align*} \sum_{\substack{G\in L(M), G\subset F}}S(G)=\sum_{A\subset F}(-1)^{|A|}. \end{align*} The right-hand side is $(1-1)^{|F|}$. Therefore it is $1$ when $F=\varnothing$ and $0$ when $F\ne\varnothing$. By [citetheorem:8094] on the finite poset $L(M)$, equivalently by the defining recursion for the Möbius function, this gives \begin{align*} S(F)=\mu_{L(M)}(\varnothing,F) \end{align*} for every flat $F\in L(M)$. Since $\rho(A)=\rho(\operatorname{cl}_M(A))$, we obtain \begin{align*} \chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}. \end{align*}[/step]

[guided]The characteristic polynomial of $L(M)$ is defined by summing the Möbius function over flats: \begin{align*} \chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}. \end{align*} To rewrite this as a sum over all subsets of $E$, we group subsets according to their closure. Define the map \begin{align*} S:L(M)&\to \mathbb Z \end{align*} \begin{align*} F&\mapsto \sum_{\substack{A\subset E, \operatorname{cl}_M(A)=F}}(-1)^{|A|}. \end{align*} We want to prove that $S(F)$ is exactly $\mu_{L(M)}(\varnothing,F)$. Fix a flat $F$. A subset $A\subset E$ has closure contained in $F$ if and only if $A\subset F$, because $F$ is closed. Therefore, summing $S(G)$ over all flats $G\subset F$ counts every subset $A\subset F$ exactly once, namely under the flat $G=\operatorname{cl}_M(A)$. Thus \begin{align*} \sum_{\substack{G\in L(M), G\subset F}}S(G)=\sum_{A\subset F}(-1)^{|A|}. \end{align*} The last sum is the binomial expansion of $(1-1)^{|F|}$. It equals $1$ when $F=\varnothing$ and equals $0$ when $F\ne\varnothing$. Because $M$ is loopless, $\operatorname{cl}_M(\varnothing)=\varnothing$, so $\varnothing$ is the minimum flat. The defining recursion for the Möbius function on the finite poset $L(M)$, equivalently [citetheorem:8094] applied to the interval from $\varnothing$ to $F$, says that the unique function $F\mapsto \mu_{L(M)}(\varnothing,F)$ satisfies \begin{align*} \sum_{\substack{G\in L(M), G\subset F}}\mu_{L(M)}(\varnothing,G)=1 \end{align*} when $F=\varnothing$, and satisfies \begin{align*} \sum_{\substack{G\in L(M), G\subset F}}\mu_{L(M)}(\varnothing,G)=0 \end{align*} when $F\ne\varnothing$. The function $S$ satisfies the same equations, hence $S(F)=\mu_{L(M)}(\varnothing,F)$ for every flat $F$. Finally, rank is unchanged by closure, so $\rho(A)=\rho(\operatorname{cl}_M(A))$. Substituting the identity for $S(F)$ into the characteristic polynomial gives \begin{align*} \chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}. \end{align*}[/guided]

[step:Define the broken-circuit cancellation domain] Let $\mathcal B$ be the set of all subsets $A\subset E$ that contain a broken circuit. Thus $A\in\mathcal B$ if and only if there exists a circuit $C$ of $M$ such that \begin{align*} C\setminus\{\min_{<}C\}\subset A. \end{align*} Define the map \begin{align*} e:\mathcal B&\to E \end{align*} by letting $e(A)$ be the smallest element $e\in E$ for which there exists a circuit $C$ of $M$ satisfying \begin{align*} \min_{<}C=e \end{align*} and \begin{align*} C\setminus\{e\}\subset A. \end{align*} Such an element exists by the definition of $\mathcal B$, and it is unique because $E$ is finite and linearly ordered. Choose one circuit with this property and denote it by $C_A$. Define a map \begin{align*} \Phi:\mathcal B&\to 2^E \end{align*} by \begin{align*} \Phi(A)=A\triangle\{e(A)\}, \end{align*} where $\triangle$ denotes symmetric difference. Equivalently, $\Phi(A)$ is obtained from $A$ by toggling the single element $e(A)$. [/step]

[step:Show that toggling preserves the chosen broken-circuit witness]We claim that $\Phi(A)\in\mathcal B$ and $e(\Phi(A))=e(A)$ for every $A\in\mathcal B$. Let $e=e(A)$ and $C=C_A$. Since $e=\min_{<}C$, the element $e$ is not contained in $C\setminus\{e\}$. Hence \begin{align*} C\setminus\{e\}\subset A\triangle\{e\}=\Phi(A). \end{align*} Thus $\Phi(A)\in\mathcal B$, and $e$ is a candidate for $e(\Phi(A))$. It remains to prove that no smaller element is a candidate for $\Phi(A)$. Suppose, for contradiction, that there exist an element $d<e$ and a circuit $D$ such that $\min_{<}D=d$ and \begin{align*} D\setminus\{d\}\subset \Phi(A). \end{align*} If $e\notin D\setminus\{d\}$, then $D\setminus\{d\}\subset A$, contradicting the minimality of $e=e(A)$. Thus $e\in D$. Since $d<e=\min_{<}C$, we have $d\notin C$. The strong circuit elimination axiom, applied to the circuits $C$ and $D$, the common element $e\in C\cap D$, and the element $d\in D\setminus C$, gives a circuit $H$ satisfying \begin{align*} d\in H\subset (C\cup D)\setminus\{e\}. \end{align*} Every element of $C\setminus\{e\}$ lies in $A$, and every element of $D\setminus\{d,e\}$ lies in $A$. Therefore \begin{align*} H\setminus\{d\}\subset A. \end{align*} Moreover $d=\min_{<}H$, because all elements of $C\setminus\{e\}$ are greater than $e$ and all elements of $D\setminus\{d\}$ are greater than $d$. Hence $d$ is a candidate for $e(A)$, contradicting $d<e(A)$. Therefore no candidate smaller than $e$ exists for $\Phi(A)$, and so \begin{align*} e(\Phi(A))=e(A). \end{align*}[/step]

[guided]This is the delicate point of the proof. We define the involution by toggling the smallest possible first element of a broken-circuit witness, and we must prove that this smallest element does not change after the toggle. Fix $A\in\mathcal B$. Write $e=e(A)$, and choose a circuit $C=C_A$ such that $\min_{<}C=e$ and \begin{align*} C\setminus\{e\}\subset A. \end{align*} The set $C\setminus\{e\}$ does not contain $e$, so toggling $e$ does not affect this containment. Hence \begin{align*} C\setminus\{e\}\subset A\triangle\{e\}=\Phi(A). \end{align*} This proves that $\Phi(A)$ still contains a broken circuit, and that $e$ is still a possible choice for the first element of a witnessing circuit. Now suppose that after toggling, a smaller element $d<e$ became possible. Then there is a circuit $D$ with $\min_{<}D=d$ and \begin{align*} D\setminus\{d\}\subset \Phi(A). \end{align*} If $e$ is not in $D\setminus\{d\}$, then toggling $e$ did not affect the set $D\setminus\{d\}$, so we would have \begin{align*} D\setminus\{d\}\subset A. \end{align*} This would make $d$ a valid candidate for $e(A)$, contradicting the minimality of $e(A)=e$. Therefore the only possible way a new smaller candidate could appear is if $e\in D$. We then use circuit elimination. The circuits $C$ and $D$ both contain $e$, and $d\in D\setminus C$ because $d<e=\min_{<}C$. The strong circuit elimination axiom gives a circuit $H$ such that \begin{align*} d\in H\subset (C\cup D)\setminus\{e\}. \end{align*} The elements of $C\setminus\{e\}$ already lie in $A$, and the elements of $D\setminus\{d,e\}$ also lie in $A$ because they are not the toggled element. Hence all elements of $H$ except possibly $d$ lie in $A$: \begin{align*} H\setminus\{d\}\subset A. \end{align*} Since $d$ is the least element of $D$ and every element of $C\setminus\{e\}$ is greater than $e>d$, we also have $\min_{<}H=d$. Thus $d$ was already a valid candidate for $e(A)$ before the toggle, again contradicting $d<e(A)$. This contradiction proves that no smaller candidate appears after toggling. Since $e$ remains a candidate, we conclude \begin{align*} e(\Phi(A))=e(A). \end{align*}[/guided]

[step:Use the toggle map as a rank-preserving sign-reversing involution] For $A\in\mathcal B$, let $e=e(A)$ and choose $C=C_A$ as above. Since \begin{align*} C\setminus\{e\}\subset A \end{align*} and $C$ is a circuit, the element $e$ lies in the closure of $C\setminus\{e\}$, hence also in the closure of $A\setminus\{e\}$ whenever $e\in A$, and in the closure of $A$ whenever $e\notin A$. Therefore toggling $e$ does not change rank: \begin{align*} \rho(\Phi(A))=\rho(A). \end{align*} The previous step gives $e(\Phi(A))=e(A)$, so toggling once more returns to $A$: \begin{align*} \Phi(\Phi(A))=A. \end{align*} Also, $A$ and $\Phi(A)$ differ by exactly one element, so \begin{align*} (-1)^{|\Phi(A)|}=-(-1)^{|A|}. \end{align*} Thus $\Phi$ is a rank-preserving sign-reversing involution on $\mathcal B$. Consequently, \begin{align*} \sum_{A\in\mathcal B}(-1)^{|A|}t^{r-\rho(A)}=0. \end{align*} [/step]

[step:Identify the surviving subsets and compare coefficients] Let $\mathcal N$ be the set of no-broken-circuit subsets of $E$. Since every subset of $E$ either lies in $\mathcal B$ or in $\mathcal N$, the subset expansion and the cancellation just proved give \begin{align*} \chi_{L(M)}(t)=\sum_{A\in\mathcal N}(-1)^{|A|}t^{r-\rho(A)}. \end{align*} Every no-broken-circuit subset is independent. Indeed, if $A\in\mathcal N$ were dependent, then $A$ would contain a circuit $C$, and therefore it would contain the broken circuit $C\setminus\{\min_{<}C\}$, contradicting the definition of $\mathcal N$. Thus $\rho(A)=|A|$ for every $A\in\mathcal N$, and so \begin{align*} \chi_{L(M)}(t)=\sum_{A\in\mathcal N}(-1)^{|A|}t^{r-|A|}. \end{align*} Grouping this sum by cardinality gives \begin{align*} \chi_{L(M)}(t)=\sum_{i=0}^{r}(-1)^i |\{A\in\mathcal N:|A|=i\}|t^{r-i}. \end{align*} Comparing this expression with \begin{align*} \chi_{L(M)}(t)=\sum_{i=0}^{r}(-1)^i w_i t^{r-i} \end{align*} shows that \begin{align*} w_i=|\{A\subset E:A\text{ is no-broken-circuit and }|A|=i\}| \end{align*} for every $0\le i\le r$. This is the desired coefficient formula. [/step]