I asked this question on Stack Exchange two weeks ago, and didn't get any answers, so I'm shamelessly reposting it here.

Let $S=\{v_1,\dots, v_n\} \subset V$ be a set of nonzero vectors in a vector space (we can take the field to be algebraically closed). It is easy to prove that if there exists a sequence $S_1,\dots, S_r \subseteq S$ such that

  1. $\dim\text{span}S_j=1$ for all $j=1,\dots, r$.

  2. $S_j \cap S_{j+1} \neq \emptyset$ for all $j=1,\dots, r-1$.

  3. $\bigcup_{j=1}^r S_j=S$.

then $\dim\text{span}S=1$.

Essentially, this gives a way to upper bound $\dim\text{span}S$ in terms of $\dim\text{span}$ of subsets obeying certain properties. My open-ended question is whether there are "interesting" generalizations of this to the case in which we want to upper bound $\dim\text{span}S\leq d$ for some $d\geq 1$, in terms of subsets obeying certain properties.

Of course, there are trivial results in this vein: If every subset of size $\leq n-1$ has dimension $\leq d$ for some $d<n-1$, then $\dim\text{span}S\leq d$. I am looking for something more useful, that hopefully looks something like the above result for $d=1$.

  • 2
    $\begingroup$ I've included a link to avoid possible duplication of effort. It would be good to do the same thing in the other direction. $\endgroup$ Commented Feb 25, 2020 at 1:01
  • $\begingroup$ Done. Thank you $\endgroup$
    – Ben
    Commented Feb 25, 2020 at 1:03
  • $\begingroup$ If condition 1 is replace by dim span $S_j = 2$ for $j=1,\ldots,r$, then dim span $S$ can be anything from 2 to $r$. $\endgroup$ Commented Feb 25, 2020 at 1:53
  • $\begingroup$ That’s a nice observation, thanks! $\endgroup$
    – Ben
    Commented Feb 26, 2020 at 0:13

1 Answer 1


Let $S = \{v_1,...,v_n \} \subseteq V$ be a set of nonzero vectors in a vector space over an algebraically closed field and suppose that we have $S_1,...,S_r \subseteq S$ such that $S = \bigcup_{i=1}^r S_i$.

For each $S_i$, set $d_i = \mathrm{dim} \mathrm{span} (S_i)$ and $c_i = \dim \mathrm{span} (S_i \cap S_{i+1})$, (we omit $c_r$). Then we have $$ \mathrm{dim} \mathrm{span}(S) = \mathrm{dim} \mathrm{span} \left( \bigcup_{i=1}^r S_i \right) = \sum_{i=1}^r \mathrm{dim} \mathrm{span}(S_i) - \sum_{1 \leq i <j \leq r} \mathrm{dim} \mathrm{span} (S_i \cap S_j) \leq \sum_{i=1}^r \mathrm{dim} \mathrm{span}(S_i) - \sum_{i = 1}^{r-1} \mathrm{dim} \mathrm{span} (S_i \cap S_{i+1}) = \sum_{i=1}^r d_i - \sum_{i=1}^{r-1} c_i $$ Therefore, $$ \dim \mathrm{span}(S) \leq \sum_{i=1}^r d_i - \sum_{i=1}^{r-1} c_i. $$

At this point, we note that $d_i \geq c_i$ and $|S_i \cap S_{i+1}| \geq c_i$. The above gives us an upper-bound for the dimension of $S$, even if it's not very good. The special case is your original theorem: let $d_i = 1$ for all $i$ and let $S_i \cap S_{i+1} \neq \emptyset$. By the assumption that our vectors are non-zero, $\dim \mathrm{span}(S_i \cap S_{i+1}) \geq 1$. So $c_i \geq 1$ and $c_i \leq d_i$, so $c_i = 1$. Then $\dim \mathrm{span}(S) = 1$ immediately follows.

In the comment left by Brendan McKay, we have $d_i = 2$ and $1 \leq c_i \leq 2$, so $\dim \mathrm{span}(S)$ is between $r$ and $2$ inclusive.

Hopefully, this is a generalization you're satisfied with even if in general, it gives a huge range.

  • 1
    $\begingroup$ Thanks. Your bound is correct but your proof is not. The set S=S_1=S_2=S_3={e_1} contradicts your second equality. $\endgroup$
    – Ben
    Commented Feb 27, 2020 at 17:08
  • $\begingroup$ Ah, I see the error now. So right now this only works if every triple intersection is the empty set. When I'm at my computer I'll try and fix the proof, thanks! $\endgroup$ Commented Feb 27, 2020 at 21:09

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