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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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What are you assuming about the base field? Clearly this is possible for finite fields. –  Grumpy Parsnip May 16 '12 at 14:03
    
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 $V_3 = \mathcal{L}(e_1-e_2)$. –  Omar Antolín-Camarena May 16 '12 at 14:04
    
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 $V$.) –  Omar Antolín-Camarena May 16 '12 at 14:08
    
@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... –  rhenskyyy May 16 '12 at 14:13
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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. –  anon May 16 '12 at 14:15

6 Answers 6

up vote 15 down vote accepted

[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 $V$. 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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3  
+1, because this is a thinly disguised algebraic geometrey proof, using the rational normal curve $S$ :-) –  Georges Elencwajg May 16 '12 at 16:34
    
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 :-) –  Jyrki Lahtonen May 16 '12 at 16:51
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this is a fantastic proof. –  mt_ May 16 '12 at 17:00
    
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/taste 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. –  Jyrki Lahtonen May 25 '12 at 12:31

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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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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+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. –  Simon Markett May 16 '12 at 15:13
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@Simon: I have added a reference to Jacobson's Basic Algebra, a standard undergraduate book. –  Georges Elencwajg May 16 '12 at 15:41
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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. –  Simon Markett May 16 '12 at 15:46
    
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. –  Georges Elencwajg May 16 '12 at 16:02
    
I think you are right about the editing. And probably I should just stop complaining about a perfectly good solution. One last thing: Yes the fact you quote is indeed elementary, but you used fancier words in your answer. –  Simon Markett May 16 '12 at 16:08

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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Looks like "my" proof was also there... and I didn't even copy ;) –  Simon Markett May 16 '12 at 14:53
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@Simon: yes; I believe you; and I upvoted your answer. :) –  Pete L. Clark May 16 '12 at 14:57

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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+1: you are warming the heart of any algebraic geometer! The proof is indeed in Fulton's book, Chapter 1, §5 , Proposition 1. –  Georges Elencwajg May 16 '12 at 16:06

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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