My textbook asks this:

Suppose that $K$ is a finite field with $k$ elements, and that $V$ is an $r$-dimensional vector space over $K$. Show that if $V = \bigcup_{i=1}^n U_i$, where $U_1,\dotsc,U_n$ are proper subspaces of $V$, then $n\geq (k^r - 1)/(k-1)$.

Struggling to prove this for a while, I did some googling and found this paper which claims to show $n = k+1$ is possible, a result which is independent of the dimension of $V$. Which is correct?

  • 1
    $\begingroup$ Possibly related(?): math.stackexchange.com/q/60698/264 $\endgroup$ Jun 23 '13 at 8:38
  • 1
    $\begingroup$ @ZevChonoles In summary: your link shows $n\geq k+1$, my link shows that this is sharp, but my textbook says $n\geq (k^r - 1)/(k-1)$. So the job now is to verify what is proved in my link is correct... $\endgroup$
    – user71815
    Jun 23 '13 at 8:46
  • $\begingroup$ @user71815 : you sure that it is not $n\leq(k^r-1)/(k-1)$ instead? If you take a set of representatives of $\Bbb P(V)$ and for each of them consider the line spanned by it, the set of these lines covers $V$ and there are exactly $(k^r-1)/(k-1)$ of them. $\endgroup$ Jun 23 '13 at 9:13
  • 1
    $\begingroup$ Your textbook is wrong except in the case $r=2$. The general case reduces to the case $r=2$, because we can "add the remaining stuff" to all the subspaces $U_i$, i.e. we can assume that they all share an $(r-2)$ dimensional subspace $W$, and work with $V/W$. $\endgroup$ Jun 23 '13 at 10:44

Consider the linear equations $$\tag{eq} \begin{cases} x_{1} - \lambda_{j} x_{r} = 0\\ x_{r} = 0. \end{cases} $$ Here $K = \{\lambda_{1}, \dots, \lambda_{k} \}$, so there are $k+1$ equations in (eq), which define $k+1$ maximal subspaces $W_{j}$ of $V$.

Let $v \in V$. If the $r$-th component $v_{r}$ of $v$ is zero, then $v \in W_{k+1}$. If $v_{r} \ne 0$, then $v_{1} = \lambda_{j} v_{r}$ for $\lambda_{j} = v_{1} v_{r}^{-1}$, so $v \in W_{j}$. Thus $V = \bigcup_{j=1}^{k+1} W_{j}$.

So it can indeed be done with $k+1$ subspaces. The answer mentioned above proves that you cannot do better than this.

  • $\begingroup$ I will check your article, but note that my argument is an explicit version of that of @JyrkiLahtonen in his comment above. $\endgroup$ Jun 24 '13 at 7:24
  • $\begingroup$ @user71815, also, the main point of your article is that this decomposition is unique up to automorphisms. $\endgroup$ Jun 24 '13 at 8:45
  • $\begingroup$ @user71815, and then my proof is essentially the same as in your article. $\endgroup$ Jun 24 '13 at 9:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.