 Solution

Let $D$ be an effective divisor linearly equivalent to $C$. If $C$ were not a component of $D$, all intersection multiplicities between its components and $C$ would be nonnegative, contradicting $D\cdot C=C^2<0$. Hence $D=C+D^{\prime }$ with $D^{\prime }$ effective and linearly equivalent to zero. For an ample divisor $H$, this gives $H\cdot D^{\prime }=0$. Positivity of $H$ on every irreducible component of a nonzero effective divisor forces $D^{\prime }=0$. Therefore $\boxed{D=C}$, illustrating the rigidity of a negative curve.