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.)
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 $V=\cup_{i=2}^{n}\ker T_i$.