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Show that if $V$ is a vector space over an infinite field $\mathbb{F}$, then $V$ cannot be written as set-theoretic union of a finite number of proper subspaces.

levap
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user219190
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    Is your original vector space finite-dimensional? – user2566092 Feb 25 '15 at 23:33
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    It would be helpful for the people who would want to help out for them to know what your trouble is and what has gone wrong with your attempts –  Mar 07 '15 at 03:15

3 Answers3

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By contraposition: if $V$ is the set-theoretic union of $n$ proper subspaces $W_i\,$ ($1\le i\le n$), then $\lvert\mkern2mu F\,\rvert\le n-1$.

Proof. We may suppose no $W_i$ is contained in the union of the other subspaces. Let $u\in W_i,\enspace u\notin \bigcup\limits_{j\neq i}W_j$ and $v\notin W_i$.

Then $(v+Fu)\cap W_i=\varnothing$ and $(v+Fu)\cap W_j\enspace(j\neq i)$ contains at most one vector since otherwise $W_j$ would contain $u$. Hence $$\lvert\mkern2mu v+ Fu\, \rvert=\lvert\mkern2mu F\,\rvert\le n-1.$$

Corollary: Avoidance lemma for vector spaces.

Let $E$ be a vector space over an infinite field. If a subspace is contained in a finite union of subspaces, it is contained in one of them.

Note: There exists a similar (and better known) Avoidance lemma for prime ideals in commutative rings.

Bernard
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  • well done 'I like the proof – Learnmore May 27 '15 at 07:06
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    Elegant! I think it would be easier to understand if you frame your proof like this: "WLOG, assume $\exists,u\in W_1$ s.t. $u\notin \bigcup\limits_{i=2}^n W_i$. Pick $v\in V$ s.t. $v\notin W_1$. Now $(v+Fu)\cap W_1=\varnothing$ and for $j=2,\ldots, n$, $|(v+Fu)\cap W_j|\leq 1$. Thus, $|F|=|v+Fu|\leq n-1$". – Nothing special Sep 12 '24 at 20:49
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    The converse is true as well. If $V$ is a vector space over a finite field $F$ s.t. $\dim V\geq 2$ then $V$ can be written as union of $|F|+1$ number of proper subspaces. I've written a proof here. – Nothing special Sep 15 '24 at 11:35
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If your original vector space is finite-dimensional, say dimension $d$, then consider the vectors that lie on the power curve $(1,\alpha,\alpha^2,\ldots,\alpha^{d-1})$ where $\alpha$ is an arbitrary element in your field. Show that all of these vectors are distinct for distinct $\alpha$ in your field and that any $d$ of them generate your entire vector space if your vector space is $d$-dimensional. (Hint: Vandermonde determinant). Then if your space is a union of finitely many subspaces, one of them must contain infinitely many vectors on the power curve so...

user2566092
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  • Sorry I have not understood the concept of vectors to lie on the power curve. That $d $ tuple are entries from field. Now will you explain a bit more? Thanks in advance. – Ri-Li Aug 08 '17 at 21:50
  • @Ri-Li I don't know what the term power curve means yet the answer is understandable. It basically says that if there exist distinct $\alpha_1,\alpha_2,\ldots,\alpha_d$ in your field $\mathbb F$ then the set ${(1,\alpha_i,\alpha_i^2,\ldots,\alpha_i^{d-1}):i=1,\ldots,d}$ is a basis of $\mathbb F^d$. – Nothing special Sep 12 '24 at 20:14
  • It follows that $V=\bigcup_{i=1}^nW_i$ where each $W_i$ is a proper subspaces then each $W_i$ can contain at most $d-1$ number of vectors of the form $(1,\alpha,\alpha^2,\ldots,\alpha^{d-1})$. So, the union can contain at most $n(d-1)$ of them. However, the field in question was infinite. So, the union isn't the whole space. – Nothing special Sep 12 '24 at 20:34
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Assume $V \subset \bigcup_{i=1}^n V_i$. In the infinite-dimensional situation, reduce to the finite dimensional situation by choosing one vector $v_i$ from each $V\setminus V_i$. Then intersect all spaces with $\text{span}(v_1,\dots,v_n)$. The hypothesis are still satisfied (and even $V = \bigcup V_i$ now), but we can assume all involved spaces to be finite-dimensional.

For each $V_i$ choose a linear form (the coefficients are "normal vector") that vanishes on $V_i$. The product of those linear forms is a degree $n$ polynomial that vanishes on $V$ which is a contradiction $-$ over an infinite field the zero polynomial is the only polynomial that vanishes on all of $V$.

Thomas
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