(t-1,k-1,n-1)$.

-   Let $\text{der} ({{\mathbb{S}}})$ be a $q$-Steiner system ${{\mathbb{S}}}_q(t-1,k-1,n-1)$. The *residual $q$-Steiner system*, $\text{res}({{\mathbb{S}}})$, for a $q$-Steiner system ${{\mathbb{S}}}_q (t,k,n)$ (which might not exists), ${{\mathbb{S}}}$, is a set of distinct $k$-subspaces from ${{\mathbb{F}}}_q^{n-1}$ such that each $t$-subspace of ${{\mathbb{F}}}_q^{n-1}$ which is not contained in $\text{der} ({{\mathbb{S}}})$, is contained in exactly $q^t$ $k$-subspaces of $\text{res}({{\mathbb{S}}})$.

It should be noted that when $q \to 1$, i.e. for a Steiner system based on an $n$-set, each$t$-subset of the $(n-1)$-set which is not contained in the derived design, is contained in exactly one $k$-subset of the derived design. This is another indication that our definition for the $q$-analog of the residual design reflects the best transformation from subsets to subspaces.

It is interesting to know if there exists a system with the same properties of the residual design in which each $t$-subspace which is not contained in the derived design, is contained in exactly $\lambda$ subspaces of the residual design, where $\lambda < q^t$. It is not difficult to prove that this is not possible if $\lambda$ is not divisible by $q$ (the proof is left for the interested reader), but it is intriguing to know if $\lambda$ divisible by $q$ is possible.

