Let $V$ be a vector space, and let $V_i$ for $i=1,\ldots, n$ be non-zero subspaces of $V$. Is it possible that $V=\cup_{i=1}^n V_i?$
The underlying field is assumed to be infinite.
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1This is a classic problem. The result can only hold if one of the $V_i$ is the entire space (assuming the vector space is over an infinite field). – Jared May 18 '13 at 01:00
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I know the result for $n=2$ but not for arbitrary $n.$ Can we use induction? – Sriti May 18 '13 at 01:02
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3Not only is this a classic problem, it is definitely a duplicate. – Zev Chonoles May 18 '13 at 01:08
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please give me the link. – Sriti May 18 '13 at 01:08
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possible duplicate: http://math.stackexchange.com/questions/145869/a-finite-dimensional-vector-space-cannot-be-covered-by-finitely-many-proper-subs/145876#145876 – Jun 25 '14 at 23:21
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Indeed, it is a duplicate A finite-dimensional vector space cannot be covered by minitel many proper subspaces? – Anne Bauval Jun 16 '23 at 12:09
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Let $V$ be a vector space over an infinite field $k$. Assume $V=\cup_{i=1}^nV_i$, where $V_i$ are all proper subspaces. Let $u\in V_1$, and take any $v\in V\setminus V_1$. There are infinitely many vectors of the form $u+cv$ for $c\in k^*$, none of which are in $V_1$. Since there are infinitely many of them, and only finitely many subspaces, there must be some $V_j$ containing at least two of them. It follows that both $v$ and $u$ are in $V_j$ so that $V_1\subset \cup_{i=2}^nV_i$.
Continuing this process, we can show that $V=V_n$, contradicting the assumption that the subspaces are proper.
I guess the assumption that $k$ is infinite is a bit overkill. We only need that there are at least as many elements in the field as subspaces in the union to derive a contradiction. This shows that even over finite fields, we need at least as many subspaces as elements in the field to obtain such a decomposition.
Jared
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It was a duplicate. https://math.stackexchange.com/a/145876/386889 – Anne Bauval Jun 16 '23 at 12:09