 Solution

The three eigenvectors form an orthonormal basis, so an arbitrary vector has expansion $\mathbf{r}=a{\mathbf{v}}_{-}+b{\mathbf{v}}_0+c{\mathbf{v}}_{+}$. Multiplication by $A$ removes the zero-eigenvalue component:Its dot product with ${\mathbf{v}}_0$ is zero for every $a,b,c$. Thus one convenient answer is the nonzero vectorEquivalently, $A^{\mathsf{T}}\mathbf{u}=A\mathbf{u}=0$, so $(A\mathbf{r})\cdot \mathbf{u}=\mathbf{r}\cdot (A^{\mathsf{T}}\mathbf{u})=0$. Any nonzero scalar multiple of this vector in the null space also works.