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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.

Martin Sleziak
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rhenskyyy
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    What are you assuming about the base field? Clearly this is possible for finite fields. – Cheerful Parsnip May 16 '12 at 14:03
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    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
  • @anon Could you give me some examples about finite fields and vector spaces over them? Thanks a lot~~ – rhenskyyy May 16 '12 at 14:18
  • Are you familiar with modular arithmetic? At any rate, a simple example would be e.g. two-dimensional space over $\Bbb F_2$, the field with two elements - namely $0$ and $1$, w/ the arithmetic rules $$0\cdot0=0\cdot1=0,~1\cdot1=1,~0+0=1+1=0,~0+1=1,$$ which you can check satisfies all the field axioms, and the only four vectors then are $(0,0),(0,1),(1,0),(1,1)$ with the usual componentwise addition and scalar multiplication. The three one-dimensional subspaces are generated by the three nonzero vectors, and in each one of these the only two elements are the zero and the generating vector. – anon May 16 '12 at 14:25
  • are you thinking of a direct sum or a union? – Rajesh D May 16 '12 at 15:15
  • See this previous question. – Arturo Magidin May 16 '12 at 19:04

6 Answers6

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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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Jyrki Lahtonen
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    +1, because this is a thinly disguised algebraic geometry proof, using the rational normal curve $S$ :-) – Georges Elencwajg May 16 '12 at 16:34
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    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. – Matthew Towers 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 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
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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.

Simon Markett
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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.

Georges Elencwajg
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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
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    @Georges: another reference for the result about polynomials versus polynomial functions you cite is Theorem 167 in http://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...) – Pete L. Clark May 16 '12 at 16:44
  • Dear @Pete, thanks for the link to your notes. There is no date but they seem to be much more complete than what I remember: is this a new version? As for Jacobson, you know of course infinitely better than I (or is it "than me"?) what is or isn't used in the USA. I had the impression that it was used because it is often mentioned in bibliographies. Does your enigmatic last word "but..." mean that it is now considered too difficult or that it no longer corresponds to the standard curriculum? – Georges Elencwajg May 16 '12 at 16:59
  • @Georges: as far as I know, BAI is no longer considered at all, but yes, I assume that at some point before my time people decided it was too difficult for undergraduates. As for the curriculum: the first four chapters have almost everything I would like to see in a year-long honors undergraduate algebra sequence...plus a little more. 300 pages of text for a year's worth of hard work is a bit compressed by American standards. The last four chapters contain stuff -- again, wonderful stuff! -- that is well beyond what one learns in standard undergraduate or graduate algebra courses. – Pete L. Clark May 16 '12 at 17:31
  • (When I wrote that 300 pages of text for a serious year-long course is a bit compressed, I couldn't help but think of http://math.uga.edu/~pete/2400full.pdf, the lecture notes from a course I just finished for first year undergraduates. But my approach to this material scared off some quite strong students, which is really not the American way, and I am not so comfortable with that aspect of it myself.) – Pete L. Clark May 16 '12 at 17:37
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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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Pete L. Clark
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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).

Makoto Kato
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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).

wildildildlife
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