 Solution

For a symmetric matrix,
$\mathrm{range}V=(\ker V{)}^{\perp }$. Since $q\notin \mathrm{range}V$, there is
$\eta \in \ker V$ with ${\eta }^{T}q\ne 0$. For any target excess mean $h$, setThen ${\theta }^{T}q=h$ andThus the required minimum is zero for every $m,x$.

Choose the sign of $\eta$ so that ${\eta }^{T}q>0$, 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 ${\eta }^{T}q$, so it is a strictly positive constant almost surely. This is an arbitrage.