Let $d=\mu-(1+r)P_0$. Since $V$ is symmetric, $A=\operatorname{Im}V=(\ker V)^\perp$. If $d\in A$, every $H\in\ker V$ satisfies
$$
H\cdot\mu=(1+r)H\cdot P_0.
$$
Any $H\notin\ker V$ has a nondegenerate normal terminal value and cannot be nonnegative almost surely. Any $H\in\ker V$ has the displayed deterministic relation, which excludes an arbitrage because $1+r=(\eta\cdot\mu)/(\eta\cdot P_0)>0$.

Conversely, if $d\notin A$, choose $h\in\ker V$ with $h\cdot d>0$. The zero-cost portfolio
$$
H=h-\frac{h\cdot P_0}{\eta\cdot P_0}\eta
$$
has deterministic terminal payoff
$$
H\cdot P_1=h\cdot\mu-(1+r)h\cdot P_0=h\cdot d>0.
$$
It is a terminal-consumption arbitrage. Thus no arbitrage is equivalent to $d\in A$, the <Arbitrage in a one-period Gaussian market> criterion.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 211 / 1 / e / Solution

{c}

If the terminal price vector is normal with covariance matrix $V$, every portfolio outside $\ker V$ has a terminal value with full support on the real line. Arbitrage can therefore arise only from portfolios in $\ker V$, whose terminal values are deterministic.


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 Incoming links: Arbitrage in a one-period Gaussian market