Union of proper subspaces, which is correct?

Question

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?

Possibly related(?): http://math.stackexchange.com/q/60698/264 — Zev Chonoles, Jun 23 '13 at 08:38
@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... — , Jun 23 '13 at 08:46
@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. — Andrea Mori, Jun 23 '13 at 09:13
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 $V/W$. — Jyrki Lahtonen, Jun 23 '13 at 10:44

score 2 · Accepted Answer · edited Apr 13 '17 at 12:20

2

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.

edited Apr 13 '17 at 12:20

Community

1

answered Jun 23 '13 at 10:23

Andreas Caranti

68,873

I will check your article, but note that my argument is an explicit version of that of @JyrkiLahtonen in his comment above. – Andreas Caranti Jun 24 '13 at 07:24
@user71815, also, the main point of your article is that this decomposition is unique up to automorphisms. – Andreas Caranti Jun 24 '13 at 08:45
@user71815, and then my proof is essentially the same as in your article. – Andreas Caranti Jun 24 '13 at 09:36

Union of proper subspaces, which is correct?

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