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Let $V$ and $W$ be vector spaces and $T_1$, $T_2$, $\dots$, $T_n$ be linear transformations from $V$ to $W$, such that for every $v$ in $V$, either $T_1 v = 0$, $T_2 v = 0$, $\dots$ or $T_n v = 0$. Can we conclude that $T_1 = 0$, $T_2 = 0$, $\dots$ or $T_n = 0$?

My attempt:

I could prove the statement for $n = 2$ but I couldn't generalize it, and I also couldn't find any counterexample for greater $n$.

For $n = 2$, suppose $T_1 \ne 0$ and $T_2 \ne 0$. Hence there are $u$ and $v$ in $V$ such that $T_1 u \ne 0$ and $T_2 v \ne 0$. By hypothesis, we must have $T_1 v = 0$ and $T_2 u = 0$ which yield $T_1 ( u + v ) \ne 0$ and $T_2 ( u + v ) \ne 0$, contradicting the hypothesis.

I also noted that my argument for $n = 2$ can be reformulated if $V$ and $W$ were just groups and $T_1$ and $T_2$ were group homomorphisms. So I got interested in the more general problem about groups. (This is in fact another question about which I'm less concerned in this particular post.)

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    $\begingroup$ Are you only considering vector spaces over infinite fields (like the real or complex numbers)? Otherwise the implication does not hold. $\endgroup$
    – Servaes
    Jun 24, 2016 at 13:12
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    $\begingroup$ If $V$ is a finite-dimensional real vector space, the claim is true. Notice that the condition is equivalent to saying that $\bigcup_{i=1}^{n} \ker T_i = V$. But if none of $T_i$ is zero, then $\bigcup_{i=1}^{n} \ker T_i$ is a finite union of nowhere dense set and hence cannot cover $V$. Of course this argument extends to a more general setting. On the other hand, if $V$ is a finite dimensional vector space over a finite field, then certainly the implication is false. $\endgroup$ Jun 24, 2016 at 13:12
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    $\begingroup$ @Sangchul Lee That seems like a perfectly fine answer to me. $\endgroup$
    – Servaes
    Jun 24, 2016 at 13:13
  • $\begingroup$ @Servaes, Thank you. But please let me wait for an answer that does not rely on topological or measure-theoretic argument. I guess my answer is an overkill... $\endgroup$ Jun 24, 2016 at 13:16
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    $\begingroup$ One can prove that a finite union of proper subspaces of a vector space $V$ over an infinite field can't be all of $V$, without resorting to topology or measure theory. I think Pete Clark was involved in a discussion of this, maybe on MathOverflow, some years ago....found it! mathoverflow.net/questions/26/… $\endgroup$ Jun 24, 2016 at 13:27

1 Answer 1

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Let $k$ be the base field. We suppose that $n$ is such that $n+1\leq |k|$.
The hypothesis gives $V = \cup_{i=1}^{n}\ker T_i$.
We show that if $\ker T_1\ne V$ then $V=\cup_{i=2}^{n}\ker T_i$, and so step by step we get an $i$ such that $V = \ker T_i$.
Suppose $\ker T_1\ne V$. Then let $x\in \ker T_1$ and $y\in V\setminus\ker T_1$. For $a\in k^\times$ there are at least $n$ different vectors of the form $x+ay\in \cup_{i=2}^{n}\ker T_i$, so there is $i$ such that two different vectors $x+ay$ and $x+by$ are in $\ker T_i$. This shows that $y$ and also $x$ are in $\ker T_i$, therefore $\ker T_1\subseteq \cup_{i=2}^{n}\ker T_i$ and so $V=\cup_{i=2}^{n}\ker T_i$.

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  • $\begingroup$ 1) Why $n+1 \leq p$? 2) Why there are n+1 different vectors of form $x+ay$? $\endgroup$
    – Tacet
    Jun 24, 2016 at 21:13
  • $\begingroup$ $y\not =0$, so $x+my, m=1,...,p-1$ are different vectors , that is at least $n$ $\endgroup$
    – m.idaya
    Jun 24, 2016 at 22:35

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