# are intersections of kernels also kernels? [closed]

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 $\mathbb{R}^{2n}$.

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## closed as off topic by Andreas Blass, Goldstern, Fernando Muro, Will Jagy, Andres CaicedoSep 24 '12 at 5:13

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Not research level, so I'm voting to close. –  Andreas Blass Sep 23 '12 at 19:39
I don't know of any canonical way to do this. If you allow the range to be changed, then $(S, T): V \rightarrow V' \oplus V'$ works. –  Deane Yang Sep 23 '12 at 20:08
If you assume that the target space $V'$ is always the same, then $R$ may fail to exist, because $\dim V'$ may be too small. If you allow a different target space, just take a direct product $S\times T:V\toV'\times V'$. –  Sergei Ivanov Sep 23 '12 at 20:15
"Not research level" is different than "elementary". This question (phrased in terms of elementary linear algebra) is non-standard, not to be found in any textbook, and even more importantly: still unanswered! The point of the question, ie. its entire substance, is not to take a different target space. In my opinion, this question has more substance and more significance than half of the daily MO posts. At bottom, almost every problem is a problem in linear algebra. And when we find an honest unanswered linear algebra question, we should appreciate its relevance. –  J. Martel Sep 23 '12 at 21:25
Mr. Martel, I recommend you find the books by Greub on linear algebra and multilinear algebra, I believe those are separate volumes. Greub maintained a steady interest in coordinate free presentation, at the price of length of presentation. I have little interest in this myself. However, I am definitely against questions where someone, asking for help, then repeatedly says No, that one's not right. As you have such a clear idea of what you want, work on this for a few weeks, order some books on interlibrary loan. When you know something for real, ask again. –  Will Jagy Sep 23 '12 at 21:42

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 $S = - \frac{a}{b} R$).

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This doesn't directly address the motivational question, but I admit that I don't really understand the relationship between the asked question and the motivational question. Doesn't the motivational question involve thinking about sums of kernels, not intersections? –  Qiaochu Yuan Sep 23 '12 at 23:16
I don't understand how you're argument could rule out the following possibility: suppose $V=W$ has finite dimension. Then $End(V)$ is a ring. Why could we not expect that there exists a polynomial $R$ in two variables over $End(V)$ with the following mystical' powers: given any two endomorphisms $S,T$ of $V$ the polynomial $R(S,T)$ has kernel exactly equal to the intersections of the kernels of $S,T$, or equal to the sum of the kernels of $S,T$? Does your argument really annihalate this possibility? I would think that the yield $R(S,T)$ should qualify as canonical'. –  J. Martel Sep 25 '12 at 15:02
@J. Martel: it doesn't rule out that possibility. I explicitly use functoriality with respect to $V$ and $W$ separately. –  Qiaochu Yuan Sep 25 '12 at 18:09
@J. Martel: if you want to ask a question which only involves one vector space rather than two, you should probably ask it as a new question (and also clarify the relationship to your motivation, which I still don't really understand). –  Qiaochu Yuan Sep 25 '12 at 18:10

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 $f\circ R$, $f\circ S$.

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