= Solution
For a symmetric matrix,
$\operatorname{range}V=(\ker V)^\perp$. Since $q\notin\operatorname{range}V$, there is
$\eta\in\ker V$ with $\eta^Tq\ne0$. For any target excess mean $h$, set
$$
\theta=\frac{h}{\eta^Tq}\eta.
$$
Then $\theta^Tq=h$ and
$$
\operatorname{Var}(X_1)=\theta^TV\theta=0.
$$
Thus the required minimum is zero for every $m,x$.
Choose the sign of $\eta$ so that $\eta^Tq>0$, buy the risky portfolio $\eta$, and finance it by borrowing its time-zero cost in the bank. Its initial wealth is zero, while its terminal excess payoff has variance zero and positive mean $\eta^Tq$, so it is a strictly positive constant almost surely. This is an <arbitrage>.
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