[proofplan] We use the subset expansion of the Tutte polynomial and evaluate it at $(x,y)=(1-q,0)$. Multiplying by $(-1)^{r(E)}$ makes the signs simplify exactly to the alternating subset formula for the characteristic polynomial. The identity follows by comparing the two polynomial expansions term by term. [/proofplan]

[step:Evaluate the subset expansion of the Tutte polynomial at $(1-q,0)$]Let $T_M(x,y) \in \mathbb{Z}[x,y]$ denote the Tutte polynomial of $M$, written by its subset expansion: \begin{align*} T_M(x,y)=\sum_{A \subset E}(x-1)^{r(E)-r(A)}(y-1)^{|A|-r(A)}. \end{align*} Substituting $x=1-q$ and $y=0$ gives \begin{align*} T_M(1-q,0)=\sum_{A \subset E}(-q)^{r(E)-r(A)}(-1)^{|A|-r(A)}. \end{align*} Multiplying by $(-1)^{r(E)}$, the summand indexed by $A \subset E$ becomes \begin{align*} (-1)^{r(E)}(-q)^{r(E)-r(A)}(-1)^{|A|-r(A)} = (-1)^{2r(E)-2r(A)+|A|}q^{r(E)-r(A)}. \end{align*} Since $2r(E)-2r(A)$ is even, this simplifies to \begin{align*} (-1)^{r(E)}T_M(1-q,0)=\sum_{A \subset E}(-1)^{|A|}q^{r(E)-r(A)}. \end{align*}[/step]

[guided]The point of this step is to make the specialization completely explicit. The Tutte polynomial $T_M(x,y) \in \mathbb{Z}[x,y]$ is defined by summing one term over every subset $A \subset E$: \begin{align*} T_M(x,y)=\sum_{A \subset E}(x-1)^{r(E)-r(A)}(y-1)^{|A|-r(A)}. \end{align*} At the specialization $(x,y)=(1-q,0)$, the two factors become $x-1=-q$ and $y-1=-1$. Therefore \begin{align*} T_M(1-q,0)=\sum_{A \subset E}(-q)^{r(E)-r(A)}(-1)^{|A|-r(A)}. \end{align*} Now multiply the whole expression by $(-1)^{r(E)}$. For a fixed subset $A \subset E$, the sign in the corresponding summand is \begin{align*} (-1)^{r(E)}(-1)^{r(E)-r(A)}(-1)^{|A|-r(A)} = (-1)^{2r(E)-2r(A)+|A|}. \end{align*} The integer $2r(E)-2r(A)$ is even, so it contributes no sign. Hence the sign reduces to $(-1)^{|A|}$, and the power of $q$ is unchanged: \begin{align*} (-1)^{r(E)}T_M(1-q,0)=\sum_{A \subset E}(-1)^{|A|}q^{r(E)-r(A)}. \end{align*} This is the exact alternating subset sum that appears in the characteristic polynomial of a matroid.[/guided]

[step:Identify the resulting subset sum with the characteristic polynomial] By definition, the characteristic polynomial $\chi_M(q) \in \mathbb{Z}[q]$ of the finite matroid $M$ is \begin{align*} \chi_M(q)=\sum_{A \subset E}(-1)^{|A|}q^{r(E)-r(A)}. \end{align*} The expression obtained in the previous step is therefore precisely $\chi_M(q)$. Hence \begin{align*} \chi_M(q)=(-1)^{r(E)}T_M(1-q,0). \end{align*} This proves the claimed polynomial identity. [/step]