 Solution

Let $S=\mathrm{supp}x$ be the support of a vector $x$, with $|S|\le s$. Suppose the null space property holds. Every other feasible vector is $z=x+v$, where $0\ne v\in \ker A$. Splitting its L1 norm over $S$ and $S^c$ and using the triangle inequality givesThus the sparse vector $x$ is the unique minimizer in basis pursuit. Notice that the argument works for complex coordinates: it uses the absolute value inequality, rather than a real sign function.

Conversely, suppose basis pursuit uniquely recovers every sparse vector of order $s$. Fix $0\ne v\in \ker A$ and any $S$ with $|S|\le s$. Take $x=-v_S$ and $z=v_{S^c}$. These vectors have the same measurements, because $A{v}_S+A{v}_{S^c}=0$, and they are distinct since $z-x=v\ne 0$. The vector $x$ has at most $s$ nonzero coordinates, so uniqueness givesThis is the null space property for every such $S$. Uniform unique recovery by basis pursuit is equivalent to the order-$s$ null space property.