are intersections of kernels also kernels? [closed]

Question

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 $ker D_\zeta$.

Our motivation: An answer to the above question should provide a coordinate-free criterion for whether the factors of a collection of primitive $n$-vectors $\xi_1, \xi_2, \ldots $ span $\mathbb{R}^{2n}$.

Answer 1

There is no such choice, if by "canonical" you mean natural in the category-theoretic sense. I am going to rename $V'$ to $W$.

You want to find a natural map $\text{Hom}(V \oplus V, W) \to \text{Hom}(V, W)$ with certain properties. Fixing $V$ and using only naturality in $W$, it follows by the Yoneda lemma that this is equivalent to finding a map $V \to V \oplus V$ and then applying $\text{Hom}(-, W)$. This is equivalent to specifying two maps $f, g : V \to V$ and considering the map $v \mapsto f(v) \oplus g(v)$.

The only maps $V \to V$ which are natural in $V$ are given by scalar multiplication; this is a corollary of the fact that the center of $\text{GL}(V)$ is given by scalars, but this only uses naturality with respect to automorphisms and the proof is easier and more general if you make more thorough use of naturality (see this blog post).

Anyway, the upshot is that the only natural maps from pairs of linear transformations $V \to W$ to linear transformations $V \to W$ are given by taking linear combinations; in other words, all you can do is take

$$(R, S) \mapsto a R + b S$$

for some fixed scalars $a, b$. It's not hard to show that no such choice does what you want (clearly both $a$ and $b$ must be nonzero and then you can take $S = - \frac{a}{b} R$).

Answer 2

A possible answer to your general question: If you assume that "canonical" means in particular that $f \circ R(S,T) = R(f\circ S, f\circ T)$ whenever $f:V'\to V''$ is 1-1 linear, then there is no canonical solution.

Proof: If $S$ and $T$ map onto the same $k$-dimensional space $W$, but $R(S,T)$ has to have larger range, then $f$ can be defined quite arbitrarily outside $W$, without influencing $f\circ R$, $f\circ S$.