I am stuck at this statement in a book of linear algebra. Even though the author casually mentions it, I am having a hard time coming up with a proof, why it must be true. Can you guys help? I can think of a few cases where it is true. Like given a finite no of lines through centre in $\mathbb{R}^2$, we can always find a point, which doesn't lie on any of them and so on. But how to generalize it?


1 Answer 1


@paul garrett gave a counterexample for finite fields. For infinite fields, let $V_1, ..., V_k$ be proper subspaces of a vector space $V$, and let $U=\bigcup_{i=1}^kV_k\subset V$.

Now, there are two cases :

  1. If $U$ is not a vector space, then it must be a proper subset of $V$, and thus $V\setminus U$ is nonempty.
  2. If $U$ is a vector space, then, by this property (extended to a finite union of subspaces) there must be an $i\in\left\{1,...,k\right\}$ such that $V_i$ contains $V_1, ..., V_k$. But then $U=V_i$ so $V\setminus U$ is nonempty.
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    $\begingroup$ This does use the fact/assumption that the field of scalars is infinite (like $\mathbb R$, $\mathbb Q$, or $\mathbb C$), as opposed to finite fields $\mathbb F_p\approx \mathbb Z/p$, for example. $\endgroup$ Jul 2, 2020 at 18:28
  • $\begingroup$ Is the property true for finite fields? I am not sure that it is. $\endgroup$
    – charlus
    Jul 2, 2020 at 18:30
  • $\begingroup$ Can you help me with an example , where in the finite field case this assumption might break down? $\endgroup$
    – Bhaswat
    Jul 2, 2020 at 18:31
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    $\begingroup$ Proceed by induction. It is clear that the property is holds for $k=2$. Now, assuming that it is true for $k\geq 2$, we will show that it is also true for $k+1$. Let $V_1, ..., V_{k+1}$ be subspaces of a vector space $V$ and assume that $U=\bigcup_{i=1}^{k+1}V_i$ is a vector space. Remark that $U=V_{k+1} \cup\bigcup_{i=1}^k V_i$ so either $\bigcup_{i=1}^k V_i$ is a subset of $V_{k+1}$, either $V_{k+1}$ is a subset of $\bigcup_{i=1}^k V_i$. If we are in the first case, we are finished. If we are in the second case, then $U=\bigcup_{i=1}^k V_i$ is a vector space and we can apply the property. $\endgroup$
    – charlus
    Jul 2, 2020 at 18:47
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    $\begingroup$ Over a finite field such as $\mathbb F_p$, a two-dimensional vector space $V$ is (in a silly but obvious way) the finite union of lines $\mathbb F_p\cdot v$, as $v$ runs over (the finite set) $V$. So the claim does not hold in this case... but for uninteresting reasons. $\endgroup$ Jul 2, 2020 at 18:48

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