This may or may not be a totally stupid question. So let's see: 

Suppose $S,T$ are two linear transformations between some fixed pair $V, V'$ of finite-dimensional real linear vector spaces. Now suppose further that $S,T$ have nontrivial kernels in $V$ and that the intersection of these kernels is, say, nontrivial. 

Question: what is a canonical choice of linear transformation $R=R(S,T):V \to V'$ such that $R$ has kernel exactly equal to the intersection of the kernels of $S,T$?

Remark: as application consider the question of determining whether a collection $\xi_1, \xi_2, \ldots$ of primitive $k$-vectors in $\wedge ^k \mathbb{R}^{n}$ are formed by wedges of a spanning set for $\mathbb{R}^n$. That is, do the factors of $\xi_1, \xi_2, \ldots$ span $\mathbb{R}^n$? Recall that to each element $\zeta \in \wedge^k \mathbb{R}^n$ one can assign the linear transformation $D_\zeta: \mathbb{R}^n \to \wedge^{k+1} \mathbb{R}^n $ which takes $v \mapsto \zeta \wedge v$. Then if $\zeta$ is primitive, one finds $ker D_\zeta$ coincides with the $\mathbb{R}^n$-subspace spanned by the factors of $\zeta$. Explicitly, if $\zeta=v_1 \wedge \cdots \wedge v_k$, then $\{v_1, \ldots, v_k \}$ span $ker D_\zeta$. 

An answer to the above question should provide a coordinate-free criterion for whether the factors of a collection of primitive $k$-vectors $\xi_1, \xi_2, \ldots $ span $\mathbb{R}^n$.