[proofplan]
We use three standard facts about characteristic polynomials of matroids: the [characteristic polynomial](/page/Characteristic%20Polynomial) is multiplicative under direct sums, every nonempty loopless matroid has characteristic polynomial vanishing at $1$, and a coloop splits as a direct summand $U_{1,1}$. For a disconnected matroid, multiplicativity gives a product of two characteristic polynomials, and both factors vanish at $1$, so the derivative of the product also vanishes at $1$. For a matroid with a coloop, the coloop contributes a factor $t-1$, and the remaining contracted matroid is still nonempty and loopless because $|E|\ge 2$, so the derivative at $1$ again vanishes.
[/proofplan]
[step:Record the characteristic-polynomial facts used in the proof]
We use the following standard facts about characteristic polynomials of finite matroids.
First, if $N_1$ and $N_2$ are finite matroids on disjoint ground sets and $N=N_1\oplus N_2$, then
\begin{align*}
\chi_N(t)=\chi_{N_1}(t)\chi_{N_2}(t).
\end{align*}
Second, if $N$ is a nonempty loopless finite matroid, then
\begin{align*}
\chi_N(1)=0.
\end{align*}
Third, if $e$ is a coloop of a finite matroid $N$, then $N$ decomposes as
\begin{align*}
N=U_{1,1}\oplus (N/e),
\end{align*}
where $U_{1,1}$ is the one-element rank-one matroid on $\{e\}$, and hence
\begin{align*}
\chi_N(t)=(t-1)\chi_{N/e}(t).
\end{align*}
Here $N/e$ denotes the contraction of $N$ by $e$.
[/step]
[step:Show that a nontrivial direct sum has zero beta invariant]
Assume that $M$ is disconnected. Then there exist nonempty disjoint subsets $E_1,E_2\subset E$ with $E=E_1\sqcup E_2$ and matroids $M_1$ on $E_1$ and $M_2$ on $E_2$ such that
\begin{align*}
M=M_1\oplus M_2.
\end{align*}
Since $M$ is loopless, both $M_1$ and $M_2$ are loopless. Since $E_1$ and $E_2$ are nonempty, the vanishing fact for nonempty loopless matroids gives
\begin{align*}
\chi_{M_1}(1)=0
\end{align*}
and
\begin{align*}
\chi_{M_2}(1)=0.
\end{align*}
By multiplicativity of the characteristic polynomial under direct sums,
\begin{align*}
\chi_M(t)=\chi_{M_1}(t)\chi_{M_2}(t).
\end{align*}
Differentiating this identity in $\mathbb{C}[t]$ and evaluating at $t=1$ gives
\begin{align*}
\chi_M'(1)=\chi_{M_1}'(1)\chi_{M_2}(1)+\chi_{M_1}(1)\chi_{M_2}'(1).
\end{align*}
Substituting $\chi_{M_1}(1)=0$ and $\chi_{M_2}(1)=0$ yields
\begin{align*}
\chi_M'(1)=0.
\end{align*}
Therefore, by the definition of $\beta(M)$,
\begin{align*}
\beta(M)=(-1)^{r(M)-1}\chi_M'(1)=0.
\end{align*}
[guided]
Assume that $M$ is disconnected. By the definition of disconnectedness in the statement, we may write the ground set as a disjoint union
\begin{align*}
E=E_1\sqcup E_2
\end{align*}
with both $E_1$ and $E_2$ nonempty, and we may write
\begin{align*}
M=M_1\oplus M_2,
\end{align*}
where $M_i$ is a matroid on $E_i$ for $i\in\{1,2\}$.
The key point is that the characteristic polynomial turns direct sums into products. Applying multiplicativity to this direct-sum decomposition gives
\begin{align*}
\chi_M(t)=\chi_{M_1}(t)\chi_{M_2}(t).
\end{align*}
We now check why both factors vanish at $t=1$. Since $M$ is loopless, no element of $E$ is a loop in $M$. In a direct sum, loops of $M_1$ and loops of $M_2$ are precisely loops of $M$ lying in the corresponding summand. Hence both $M_1$ and $M_2$ are loopless. Since $E_1$ and $E_2$ are nonempty, both summands are nonempty loopless finite matroids. The standard vanishing property of the characteristic polynomial for a nonempty loopless matroid therefore gives
\begin{align*}
\chi_{M_1}(1)=0
\end{align*}
and
\begin{align*}
\chi_{M_2}(1)=0.
\end{align*}
Differentiate the product identity using the ordinary product rule for polynomials over $\mathbb{C}$:
\begin{align*}
\chi_M'(t)=\chi_{M_1}'(t)\chi_{M_2}(t)+\chi_{M_1}(t)\chi_{M_2}'(t).
\end{align*}
Evaluating at $t=1$ gives
\begin{align*}
\chi_M'(1)=\chi_{M_1}'(1)\chi_{M_2}(1)+\chi_{M_1}(1)\chi_{M_2}'(1).
\end{align*}
Each summand on the right contains one of the values $\chi_{M_1}(1)$ or $\chi_{M_2}(1)$, and both of those values are zero. Thus
\begin{align*}
\chi_M'(1)=0.
\end{align*}
Finally, the beta invariant is defined by
\begin{align*}
\beta(M)=(-1)^{r(M)-1}\chi_M'(1).
\end{align*}
Since $\chi_M'(1)=0$, multiplication by the sign $(-1)^{r(M)-1}$ still gives
\begin{align*}
\beta(M)=0.
\end{align*}
[/guided]
[/step]
[step:Show that a coloop gives a double vanishing at $t=1$]
Assume that $M$ has a coloop $e\in E$. Since $|E|\ge 2$, the set $E\setminus\{e\}$ is nonempty. Since $M$ is loopless and $e$ is a coloop, the contraction $M/e$ is a loopless matroid on the nonempty ground set $E\setminus\{e\}$.
By the coloop factorization of the characteristic polynomial,
\begin{align*}
\chi_M(t)=(t-1)\chi_{M/e}(t).
\end{align*}
Differentiating gives
\begin{align*}
\chi_M'(t)=\chi_{M/e}(t)+(t-1)\chi_{M/e}'(t).
\end{align*}
Evaluating at $t=1$ gives
\begin{align*}
\chi_M'(1)=\chi_{M/e}(1).
\end{align*}
Because $M/e$ is nonempty and loopless, the characteristic-polynomial vanishing property gives
\begin{align*}
\chi_{M/e}(1)=0.
\end{align*}
Thus
\begin{align*}
\chi_M'(1)=0.
\end{align*}
By the definition of the beta invariant,
\begin{align*}
\beta(M)=(-1)^{r(M)-1}\chi_M'(1)=0.
\end{align*}
[guided]
Assume that $M$ has a coloop $e\in E$. The hypothesis $|E|\ge 2$ is used here: after removing or contracting the one element $e$, the remaining ground set
\begin{align*}
E\setminus\{e\}
\end{align*}
is still nonempty.
Because $e$ is a coloop, it splits off as a one-element rank-one direct summand. Equivalently,
\begin{align*}
M=U_{1,1}\oplus (M/e),
\end{align*}
where $U_{1,1}$ is the one-element rank-one matroid on $\{e\}$. The characteristic polynomial of $U_{1,1}$ is $t-1$, so multiplicativity under direct sums gives the coloop factorization
\begin{align*}
\chi_M(t)=(t-1)\chi_{M/e}(t).
\end{align*}
We must also check that the remaining factor vanishes at $t=1$. The contraction $M/e$ has ground set $E\setminus\{e\}$, which is nonempty because $|E|\ge 2$. It is loopless: if an element $f\in E\setminus\{e\}$ became a loop in $M/e$, then $\{e,f\}$ would contain a circuit of $M$ involving the coloop $e$, contradicting the defining property that a coloop lies in no circuit. Since $M$ itself is loopless, no element outside $e$ was already a loop. Hence $M/e$ is a nonempty loopless finite matroid.
Now differentiate the factorization
\begin{align*}
\chi_M(t)=(t-1)\chi_{M/e}(t).
\end{align*}
Using the product rule for polynomials,
\begin{align*}
\chi_M'(t)=\chi_{M/e}(t)+(t-1)\chi_{M/e}'(t).
\end{align*}
At $t=1$, the second term contains the factor $1-1$, so
\begin{align*}
\chi_M'(1)=\chi_{M/e}(1).
\end{align*}
Since $M/e$ is nonempty and loopless, the standard vanishing property gives
\begin{align*}
\chi_{M/e}(1)=0.
\end{align*}
Therefore
\begin{align*}
\chi_M'(1)=0.
\end{align*}
Substituting this into the defining formula
\begin{align*}
\beta(M)=(-1)^{r(M)-1}\chi_M'(1)
\end{align*}
gives
\begin{align*}
\beta(M)=0.
\end{align*}
[/guided]
[/step]
[step:Conclude both asserted vanishing statements]
The disconnected case has shown that every nontrivial direct-sum decomposition of the loopless matroid $M$ forces
\begin{align*}
\beta(M)=0.
\end{align*}
The coloop case has shown that every coloop of $M$ forces
\begin{align*}
\beta(M)=0.
\end{align*}
These are exactly the two conclusions of the theorem.
[/step]
