[proofplan]
We first translate the derivative definition of $\beta(M)$ into the coefficient of $x$ in the specialization $T_M(x,0)$ of the Tutte polynomial, after treating loops separately. This coefficient description makes disconnectedness easy: the Tutte polynomial is multiplicative under direct sums, and every positive-rank loopless summand contributes zero constant term after setting $y=0$. Positivity and duality are then supplied by Crapo's activity interpretation of the beta invariant: for a connected matroid with at least two elements, $\beta(M)$ counts bases with exactly one internal active element and no external active elements, and duality exchanges internal and external activity under complementation of bases.
[/proofplan]
[step:Convert the derivative definition into a Tutte coefficient]
Let $r=r_M(E)$ denote the rank of $M$, and let $T_M(x,y)\in \mathbb{Z}[x,y]$ denote its Tutte polynomial. For a polynomial $P(x,y)\in \mathbb{Z}[x,y]$, define $[x^a y^b]P$ to be the coefficient of the monomial $x^a y^b$ in $P$. In particular,
\begin{align*}
[x]T_M(x,0)=[x^1y^0]T_M(x,y).
\end{align*}
If $M$ has a loop, then $\chi_M(q)=0$ for all $q$, so $\beta(M)=(-1)^{r-1}\chi_M'(1)=0$.
Assume now that $M$ is loopless. We use the standard characteristic-polynomial specialization of the Tutte polynomial, namely
\begin{align*}
\chi_M(q)=(-1)^r T_M(1-q,0)
\end{align*}
(citing a result not yet in the wiki: [characteristic polynomial as a Tutte polynomial specialization](/theorems/6648)). Differentiating with respect to $q$ and evaluating at $q=1$ gives
\begin{align*}
\chi_M'(1)=(-1)^{r+1}\frac{\partial T_M}{\partial x}(0,0).
\end{align*}
Therefore
\begin{align*}
\beta(M)=(-1)^{r-1}\chi_M'(1)=\frac{\partial T_M}{\partial x}(0,0)=[x]T_M(x,0).
\end{align*}
[guided]
We want to replace the derivative at $q=1$ by a coefficient of the Tutte polynomial. This is useful because disconnectedness and duality are much more transparent for $T_M(x,y)$ than for $\chi_M(q)$.
First suppose $M$ has a loop. The characteristic polynomial of a matroid with a loop is identically zero, so $\chi_M'(1)=0$. Since $r>0$, the defining formula
\begin{align*}
\beta(M)=(-1)^{r-1}\chi_M'(1)
\end{align*}
immediately gives $\beta(M)=0$.
Now suppose $M$ is loopless. The standard relation between the characteristic polynomial and the Tutte polynomial is
\begin{align*}
\chi_M(q)=(-1)^r T_M(1-q,0)
\end{align*}
(citing a result not yet in the wiki: characteristic polynomial as a Tutte polynomial specialization). The substitution is $x=1-q$ and $y=0$. Applying the ordinary chain rule to the polynomial $q\mapsto T_M(1-q,0)$ gives
\begin{align*}
\chi_M'(q)=(-1)^r\left(-\frac{\partial T_M}{\partial x}(1-q,0)\right).
\end{align*}
Evaluating at $q=1$ yields
\begin{align*}
\chi_M'(1)=(-1)^{r+1}\frac{\partial T_M}{\partial x}(0,0).
\end{align*}
Multiplying by $(-1)^{r-1}$ gives
\begin{align*}
\beta(M)=(-1)^{r-1}\chi_M'(1)=\frac{\partial T_M}{\partial x}(0,0).
\end{align*}
Finally, for a polynomial in $x$ and $y$, the derivative $\frac{\partial T_M}{\partial x}(0,0)$ extracts exactly the coefficient of $x^1y^0$. Hence
\begin{align*}
\beta(M)=[x]T_M(x,0).
\end{align*}
This notation means the coefficient of the monomial $x^1y^0$ in $T_M(x,y)$, not the sum of all terms of $x$-degree one.
[/guided]
[/step]
[step:Use multiplicativity of the Tutte polynomial to prove vanishing for disconnected matroids]
Assume $M$ is disconnected. If $M$ has a loop, the preceding step already gives $\beta(M)=0$. Hence assume $M$ is loopless.
Since $M$ is disconnected, its ground set decomposes as a disjoint union of at least two nonempty connected components
\begin{align*}
E=E_1\sqcup \cdots \sqcup E_m
\end{align*}
with $m\geq 2$, and
\begin{align*}
M=M_1\oplus \cdots \oplus M_m,
\end{align*}
where $M_j=M|E_j$ is the restriction of $M$ to $E_j$. Because $M$ is loopless and $E_j\neq \varnothing$, each $M_j$ has positive rank.
The Tutte polynomial is multiplicative under direct sums:
\begin{align*}
T_M(x,y)=\prod_{j=1}^m T_{M_j}(x,y)
\end{align*}
(citing a result not yet in the wiki: multiplicativity of the Tutte polynomial under direct sums). For each $j$, the matroid $M_j$ is loopless and has positive rank, so the characteristic-polynomial specialization gives
\begin{align*}
T_{M_j}(0,0)=(-1)^{r_{M_j}(E_j)}\chi_{M_j}(1).
\end{align*}
The characteristic polynomial of a positive-rank loopless matroid vanishes at $1$, so $\chi_{M_j}(1)=0$ (citing a result not yet in the wiki: $\chi_P(1)=0$ for positive-rank loopless matroids). Thus $T_{M_j}(0,0)=0$ for every $j$.
Set
\begin{align*}
A_j(x)=T_{M_j}(x,0)\in \mathbb{Z}[x].
\end{align*}
Each $A_j$ has zero constant term. Since $m\geq 2$, the product $\prod_{j=1}^m A_j(x)$ has no term of degree $1$ in $x$. Therefore
\begin{align*}
[x]T_M(x,0)=0.
\end{align*}
By the coefficient formula from the preceding step, $\beta(M)=0$.
[/step]
[step:Apply Crapo's activity theorem to get positivity in the connected case]
Assume $M$ is connected and $|E|\geq 2$. Then $M$ has no loops and no coloops, since a loop or coloop is a one-element connected component.
Fix a linear order $<$ on $E$. Let $\mathcal{B}(M)$ denote the set of bases of $M$. For each basis $B\in \mathcal{B}(M)$, let $i_<(B)$ denote the number of internally active elements of $B$ with respect to $<$, and let $e_<(B)$ denote the number of externally active elements of $B$ with respect to $<$. The basis-activity expansion of the Tutte polynomial states that
\begin{align*}
T_M(x,y)=\sum_{B\in \mathcal{B}(M)}x^{i_<(B)}y^{e_<(B)}
\end{align*}
(citing a result not yet in the wiki: Tutte polynomial basis-activity expansion). Hence
\begin{align*}
[x]T_M(x,0)=\#\{B\in \mathcal{B}(M): i_<(B)=1 \text{ and } e_<(B)=0\}.
\end{align*}
By Crapo's beta theorem, for every connected matroid with at least two elements, this activity class is nonempty and its cardinality is the beta invariant (citing a result not yet in the wiki: Crapo's beta theorem). Therefore
\begin{align*}
\beta(M)=[x]T_M(x,0)>0.
\end{align*}
[/step]
[step:Exchange activities under duality to prove $\beta(M)=\beta(M^*)$]
Assume $r_N(E)>0$. If $M$ is disconnected, then $N=M^*$ is also disconnected because duality preserves direct sums, and the first part gives
\begin{align*}
\beta(M)=0=\beta(N).
\end{align*}
It remains to consider the case where $M$ is connected. Since $r_M(E)>0$ and $r_N(E)>0$, the connected matroid $M$ cannot have exactly one element; hence $|E|\geq 2$. Fix a linear order $<$ on $E$. Define the complement map on bases
\begin{align*}
\Phi:\mathcal{B}(M)\to \mathcal{B}(N),\quad B\mapsto E\setminus B.
\end{align*}
This is a bijection by the definition of dual matroid bases. Under this bijection, internal activity in $M$ becomes external activity in $N$, and external activity in $M$ becomes internal activity in $N$ (citing a result not yet in the wiki: duality of internal and external activity).
By Crapo's beta theorem applied to $M$, $\beta(M)$ is the number of bases of $M$ with exactly one internal active element and no external active elements. By the same theorem applied to $N$, $\beta(N)$ is the corresponding beta activity count for $N$. The complement bijection identifies the beta activity classes of $M$ and $N$ after exchanging internal and external activity, equivalently using $T_N(x,y)=T_M(y,x)$ together with Crapo's equality of the two beta activity counts. Hence the two counted finite sets have the same cardinality, and therefore
\begin{align*}
\beta(M)=\beta(N).
\end{align*}
This proves all asserted properties of the beta invariant.
[/step]
