$||v||_0$, so $\leq d-s$: The matrix $A$ needs to have at least one entry for every entry of $v$ (otherwise it can't obtain that entry). It is also sufficient to have so many entries, as if $u_j \not= 0$ then we can set for every entry $v_i \not= 0$ the matrix entry $A_{i,j} = \frac{v_i}{u_j}$.

$||v||_0$, so $\leq d-s$: The matrix $A$ needs to have at least one entry for every entry of $v$ (otherwise it can't obtain that entry). It is also sufficient to have so many entries, as if we consider one $j$ with $u_j \not= 0$ (which exists otherwise the system has a solution (the zero matrix) only if $v_i = 0$ for all $i$) then we can set for every entry $v_i \not= 0$ the matrix entry $A_{i,j} = \frac{v_i}{u_j}$.