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Let $V$ be a finite-dimensional vector space, $V_i$ is a proper subspace of $V$ for every $1\leq i\leq m$ for some integer $m$. In my linear algebra text, I've seen a result that $V$ can never be covered by $\{V_i\}$, but I don't know how to prove it correctly. I've written down my false proof below:

First we may prove the result when $V_i$ is a codimension-1 subspace. Since $codim(V_i)=1$, we can pick a vector $e_i\in V$ s.t. $V_i\oplus\mathcal{L}(e_i)=V$, where $\mathcal{L}(v)$ is the linear subspace span by $v$. Then we choose $e=e_1+\cdots+e_m$, I want to show that none of $V_i$ contains $e$ but I failed.

Could you tell me a simple and corrected proof to this result? Ideas of proof are also welcome~

Remark: As @Jim Conant mentioned that this is possible for finite field, I assume the base field of $V$ to be a number field.

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    $\begingroup$ What are you assuming about the base field? Clearly this is possible for finite fields. $\endgroup$
    – Cheerful Parsnip
    May 16, 2012 at 14:03
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    $\begingroup$ You're proof strategy won't work, $e$ might be zero, for example: Say $V$ is two dimensional with basis $\{e_1,e_2\}$ and take $V_1 = \mathcal{L}(e_2)$, $V_2 = \mathcal{L}(e_1)$, $e_3 = -e_1-e_2$ and $V_3 = \mathcal{L}(e_1-e_2)$. $\endgroup$
    – Omar Antolín-Camarena
    May 16, 2012 at 14:04
  • $\begingroup$ As @JimConant points out this is false when the scalars form a finite field. It is true for infinite fields of scalars however and the standard proof is to show the following more general statement by induction: $V$ is not the union of $n$ proper affine subspaces. (An affine subspace of a vector space is a translate of a vector subspace, i.e., a set of the form $\{u+a : u \in U\}$ where $U$ is some vector subspace of $V$.) $\endgroup$
    – Omar Antolín-Camarena
    May 16, 2012 at 14:08
  • $\begingroup$ @JimConant My textbook always assume the base field to be a number field. Could you tell me why this is possible for finite fields? Forgive me that I haven't learn abstract algebra and I cannot imagine a field that is not a number field... $\endgroup$
    – rhenskyyy
    May 16, 2012 at 14:13
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    $\begingroup$ A finite-dimensional vector space over a finite field in fact has a finite number of vectors. Every vector is an element of some subspace, so just take collection of these, one corresponding to every vector, and we have a cover. $\endgroup$
    – anon
    May 16, 2012 at 14:15

6 Answers 6

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[Edit] This answer is contained in another answer of mine. Sorry about that. Switching to CW[/Edit]

Pick a basis and a system of coordinates $x_1,\ldots,x_n$ for $V$. WLOG assume that $n \geq 2$. As you observed, without loss of generality we can assume that the subspaces are all of codimension one, i.e. spaces of solutions of a single homogeneous equation $$ a_1x_1+a_2x_2+\cdots +a_nx_n=0 $$ in the coordinates $x_i,i=1,\ldots,n$. Therefore a single subspace will intersect the infinite set $$ S=\{(1,t,t^2,\ldots,t^{n-1})\mid t\in k\} $$ at finitely many points, because the polynomial $a_1+a_2t+a_3t^2+\cdots+a_nt^{n-1}$ has at most $n-1$ zeros.

Therefore it is impossible to cover all of $S$, hence all of $V$, with finitely many subspaces.

Note that if $k$ is uncountable, then this argument shows that we need uncountably many subspaces.

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    $\begingroup$ +1, because this is a thinly disguised algebraic geometry proof, using the rational normal curve $S$ :-) $\endgroup$
    – Georges Elencwajg
    May 16, 2012 at 16:34
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    $\begingroup$ Thanks, @Georges. When in grad school my real analysis teacher wanted to prove the impossibility of covering a real vector space with a countable collection of subspaces using Baire category theorem. We felt that an algebraic claim needs an algebraic proof :-) $\endgroup$
    – Jyrki Lahtonen
    May 16, 2012 at 16:51
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    $\begingroup$ this is a fantastic proof. $\endgroup$
    – Matthew Towers
    May 16, 2012 at 17:00
  • $\begingroup$ I had a recollection of having used this argument earlier here. Somewhat against my expectations that other question was not tagged with finite-fields, so I didn't find it. Today another similar question was asked, and joriki found my answer here. I am switching this to CW, for it is surely bad form to try and collect upvotes for the same answer in two different locations. The other answer is more comprehensive, so this will have to go. Sorry about this. $\endgroup$
    – Jyrki Lahtonen
    May 25, 2012 at 12:31
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Do you know the proof that the union of two subspaces of a vector space is a subspace if and only if one of the two subspaces is contained in the other? If the field is infinte one can come up with a similar proof for your statement.

Assume $V$ is covered by finitely many $V_i$, and assume that the cover is minimal. Then there is wlog a $v\in V_1$ which is not in any other $V_i$ and there is also wlog a $w\in V_2-V_1$. Then the vectors $av+w$ for $0\neq a\in k$ (where $k$ is the base field) are in pairwise different spaces $V_i$. Indeed if $av+w$ and $bv+w$ both are in $V_i$, then so is $(a-b)v$ which is a contradiction. Since $k$ is infinite this proves your statement.

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  • $\begingroup$ wonderful argument!!! thank you. $\endgroup$
    – 王李远
    May 12, 2017 at 6:37
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Let me prove that that if $k$ an infinite field, a finite number of hyperplanes $H_1,\ldots, H_r$ can't cover the vector space $k^n$ .

If we had $k^n=\bigcup_{i=1}^{r} H_i$ where $H_i$ is the kernel of the non-zero linear form $l^{(i)}(x_1,\ldots,x_n)=\sum_{j=1}^{n}a_j^{(i)}x_j\in (k^n)^\ast$ , the degree $r$ polynomial $P(x_1,\ldots,x_n)=\prod_{i=1}^{r} l^{(i)}(x_1,\ldots,x_n)\in k[x_1,\ldots,x_n]$ would vanish at all points of $k^n$ without being the zero polynomial .
This is well known to be impossible if $k$ is infinite: Jacobson Theorem 2.19, page 136.

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  • $\begingroup$ +1 since this is actually a nice proof. However I'm in doubt whether it has too little details to be helpful for anybody without a wider background. $\endgroup$
    – Simon Markett
    May 16, 2012 at 15:13
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    $\begingroup$ @Simon: I have added a reference to Jacobson's Basic Algebra, a standard undergraduate book. $\endgroup$
    – Georges Elencwajg
    May 16, 2012 at 15:41
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    $\begingroup$ Don't get me wrong, the proof is nice, readable, understandable and complete. Still, I have the feeling that the theorem you refer to is harder than the actual question. Also: the fact that each hyperplane is the kernel of a non-zero linear form might be hard to grasp for an undergrad. Nothing in particular is really beyond undergrad level but I imagine that people who struggle with the original question might also struggle with your answer. $\endgroup$
    – Simon Markett
    May 16, 2012 at 15:46
  • $\begingroup$ Dear @Simon: thanks for your comment. I can't really be sure what exactly a generic undergraduate on this site finds easy or not: I thought that the fact that a hyperplane has a linear equation is pretty elementary linear algebra, but I may be wrong. Anyway, if anybody, undergraduate or not, asks a question , elementary or not, about my post I will try to answer it. I think this is better than trying to modify my answer by just guessing in advance where there might be a difficulty. $\endgroup$
    – Georges Elencwajg
    May 16, 2012 at 16:02
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    $\begingroup$ @Georges: another reference for the result about polynomials versus polynomial functions you cite is Theorem 167 in math.uga.edu/~pete/4400FULL.pdf. (I must confess that I don't view Jacobson's Basic Algebra I as a "standard undergraduate text", notwithstanding what the author writes in the preface. If anyone has any personal experience with an undergraduate course in the United States that used this text, I would be very interested to know. It's a wonderful text, but...) $\endgroup$
    – Pete L. Clark
    May 16, 2012 at 16:44
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This question was asked on MathOverflow several years ago and received many answers: please see here.

One of these answers was mine. I referred to this expository note, which has since appeared in the January 2012 issue of the American Mathematical Monthly.

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    $\begingroup$ Looks like "my" proof was also there... and I didn't even copy ;) $\endgroup$
    – Simon Markett
    May 16, 2012 at 14:53
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    $\begingroup$ @Simon: yes; I believe you; and I upvoted your answer. :) $\endgroup$
    – Pete L. Clark
    May 16, 2012 at 14:57
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This is a special case of a fact that an affine space over an infinite field is irreducible. The proof can be found in most books on elementary algebraic geometry(see for example Fulton's algebraic curves).

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    $\begingroup$ +1: you are warming the heart of any algebraic geometer! The proof is indeed in Fulton's book, Chapter 1, §5 , Proposition 1. $\endgroup$
    – Georges Elencwajg
    May 16, 2012 at 16:06
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Having been thinking about functional analysis for the past week, Baire's category theorem came to mind, but unfortunately this assumes the field is $\mathbb{R}$ or $\mathbb{C}$:

A finite dimensional linear subspace is closed, and a proper linear subspace has empty interior. So by Baire, a countable union of finite dimensional linear subspaces again has empty interior; in particular it is not the whole space.

(I believe the first sentence is still true in the generality of $V$ being a topological vector space. However to apply Baire to $V$ we need it to be locally compact Hausdorff (i.e. finite dimensional, by Riesz) or completely metrizable (e.g. a Frechet or even F- space).

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